JEE Gravitation Questions

The geostationary satellite revolves in the same sense as the earth’s rotation, so its period in an inertial frame is the sidereal day:

$$T_1 = 24 \text{ h}$$ (to the accuracy required in the problem)

According to Kepler’s third law, the time period is proportional to $$r^{3/2}$$:

$$\frac{T_2}{T_1} = \left(\frac{r_2}{r_1}\right)^{3/2}$$

Given $$r_1 = 1.21\,r_2$$, we obtain

$$T_2 = T_1 \left(\frac{1}{1.21}\right)^{3/2}$$

Compute the numerical factor:
$$1.21^{3/2} = (1.21)^{1.5} \approx 1.331$$
$$\left(\frac{1}{1.21}\right)^{3/2} \approx \frac{1}{1.331} \approx 0.752$$

Hence

$$T_2 = 24 \times 0.752 \text{ h} \approx 18.0 \text{ h}$$

The second satellite moves in the opposite (retrograde) direction, so its angular velocity is opposite in sign to that of the geostationary satellite. In magnitude:

$$\omega_1 = \frac{2\pi}{T_1}, \quad \omega_2 = -\,\frac{2\pi}{T_2}$$

Their relative angular speed is

$$\omega_{\text{rel}} = \omega_1 - \omega_2 = \frac{2\pi}{T_1} + \frac{2\pi}{T_2} = 2\pi\left(\frac{1}{T_1} + \frac{1}{T_2}\right)$$

The time period with which the second satellite is seen to pass the geostationary satellite (as measured from the geostationary frame) is therefore

$$T_{\text{rel}} = \frac{2\pi}{\omega_{\text{rel}}} = \frac{1}{\dfrac{1}{T_1} + \dfrac{1}{T_2}} = \frac{T_1\,T_2}{T_1 + T_2}$$

Substituting $$T_1 = 24\text{ h}$$ and $$T_2 \approx 18.0\text{ h}$$:

$$T_{\text{rel}} = \frac{24 \times 18.0}{24 + 18.0} \text{ h} = \frac{432}{42} \text{ h} \approx 10.3 \text{ h}$$

This observed period is given in the problem as $$\dfrac{24}{p} \text{ h}$$, so

$$\frac{24}{p} = 10.3 \Longrightarrow p = \frac{24}{10.3} \approx 2.33$$

Thus the required value is

$$\boxed{p \approx 2.3 - 2.4}$$

According to Kepler’s second law, the line joining the Sun and any planet sweeps out equal areas in equal intervals of time. This statement is the same as saying that the areal velocity $$\frac{dA}{dt}$$ of the planet is constant.

In mechanics the areal velocity of a particle of mass $$m$$ moving with linear momentum $$\mathbf{p}=m\mathbf{v}$$ at an instantaneous position vector $$\mathbf{r}$$ from the origin is given by

$$\frac{dA}{dt}=\frac{1}{2m}\,|\mathbf{r}\times\mathbf{p}|$$

The quantity $$\mathbf{L}=\mathbf{r}\times\mathbf{p}$$ is the angular momentum of the particle about the origin. Hence

$$\frac{dA}{dt}=\frac{|\mathbf{L}|}{2m}$$

If a particle is subjected to a central force, the force vector $$\mathbf{F}$$ is always directed along the radius vector $$\mathbf{r}$$. Therefore the torque about the origin is

$$\boldsymbol{\tau}=\mathbf{r}\times\mathbf{F}=0$$

Zero torque implies $$\frac{d\mathbf{L}}{dt}=0$$, so the angular momentum $$\mathbf{L}$$ is conserved (constant in both magnitude and direction).

Because $$|\mathbf{L}|$$ is constant, the expression $$\frac{dA}{dt}=\frac{|\mathbf{L}|}{2m}$$ shows that the areal velocity $$\frac{dA}{dt}$$ must also be constant. This directly yields Kepler’s second law.

Thus:

  • Assertion (A) is correct: the radius vector sweeps out equal areas in equal times, so the areal velocity is constant.
  • Reason (R) is correct: in a central force field the angular momentum is conserved.
  • Reason (R) correctly explains why the areal velocity is constant, since constant angular momentum leads to constant $$\frac{dA}{dt}$$.

Therefore the appropriate choice is Option A: Both (A) and (R) are correct and (R) is the correct explanation of (A).

The object is initially at rest at a height $$h = 3R$$ above Earth’s surface.
Therefore its initial distance from Earth’s centre is
$$r_0 = R + h = R + 3R = 4R$$.

To “never return to Earth” the object must just reach infinity with zero speed. Hence we apply conservation of mechanical energy between the initial point $$r_0$$ and infinity.

Initial mechanical energy
Kinetic: $$K_i = \frac12 m v^2$$ (where $$v$$ is the required minimum speed).
Gravitational potential: $$U_i = -\frac{GMm}{r_0} = -\frac{GMm}{4R}$$.

Final mechanical energy at infinity
Kinetic: $$K_f = 0$$ (minimum condition).
Potential: $$U_f = 0$$ (by convention).

Conservation of energy gives $$K_i + U_i = K_f + U_f$$ $$\frac12 m v^2 - \frac{GMm}{4R} = 0$$.

Solve for $$v$$:

$$\frac12 m v^2 = \frac{GMm}{4R}$$

$$v^2 = \frac{2GM}{4R} = \frac{GM}{2R}$$

$$v = \sqrt{\frac{GM}{2R}}$$.

The minimum speed required is $$\sqrt{\dfrac{GM}{2R}}$$, which corresponds to Option A.

Escape velocity: v_e = √(2GM/R). For planet: M_p = M_e/8, R_p = R_e/2.

v_p = √(2G(M/8)/(R/2)) = √(2GM/(4R)) = v_e/2 = 11.2/2 = 5.6 km/s.

The correct answer is Option 3: 5.6.

Assertion (A): A simple pendulum is taken to a planet with mass $$4M_E$$ and radius $$2R_E$$. The time period remains the same.

Reason (R): The mass of the pendulum remains unchanged.

Since gravitational acceleration is given by $$g = \frac{GM}{R^2}$$, on Earth we have $$g_E = \frac{GM_E}{R_E^2}$$ and on the planet

$$g_P = \frac{G \cdot 4M_E}{(2R_E)^2} = \frac{4GM_E}{4R_E^2} = \frac{GM_E}{R_E^2} = g_E\,. $$

From this it follows that $$g_P = g_E$$. Since the time period of a simple pendulum is $$T = 2\pi\sqrt{\frac{l}{g}}$$, the equality of gravitational acceleration implies that the time period remains the same on both Earth and the planet. Thus Assertion (A) is true.

The mass of the pendulum does remain unchanged, as mass is invariant under such a transfer, so Reason (R) is true.

However, the pendulum’s time period does not depend on its mass but only on $$g$$ and $$l$$, so the unchanged mass does not explain why the time period remains the same. Therefore, although Reason (R) is true, it does not correctly explain Assertion (A).

The correct answer is Option 4: Both (A) and (R) are true but (R) is NOT the correct explanation of (A).

For a body of mass $$m$$ projected from the earth’s surface (radius $$R$$) let the minimum speed required to reach infinity with zero speed be $$v_e$$ (escape velocity).

Gravitational potential energy at the earth’s surface is $$U_s = -\dfrac{GMm}{R}$$, while at infinity it is $$U_\infty = 0$$ (this is the highest possible value of gravitational potential energy).

Using conservation of mechanical energy, the total energy at the surface must equal the total energy at infinity:

$$\dfrac{1}{2} m v_e^2 + \left(-\dfrac{GMm}{R}\right) = 0 \quad -(1)$$

Solving $$(1)$$ gives the escape velocity:
$$v_e = \sqrt{\dfrac{2GM}{R}}$$

The kinetic energy that must be supplied is therefore
$$K = \dfrac{1}{2} m v_e^2 = \dfrac{1}{2} m \left(\dfrac{2GM}{R}\right) = \dfrac{GMm}{R}$$

Using $$g = \dfrac{GM}{R^2}$$, we rewrite this as
$$K = m g R$$

Assertion A claims the required kinetic energy is $$\dfrac{1}{2} m g R$$, which is not equal to the correct value $$m g R$$. Hence Assertion A is false.

Reason R states that “the maximum potential energy of a body is zero when it is projected to infinity from the earth’s surface.”
Because gravitational potential is defined to be zero at infinity and is negative everywhere else, this statement is true.

Therefore, A is false but R is true. The correct option is Option A.

The only quantities that decide the collapse-time for three identical masses released from the corners of an equilateral triangle are

• the initial side length $$a$$,
• the mass of each sphere $$m$$,
• the universal constant of gravitation $$G$$.

Let the required time be $$T$$.  Assume a dependence of the form

$$T \propto G^{\,p}\,m^{\,q}\,a^{\,r}$$

Write dimensions (M ≡ mass, L ≡ length, T ≡ time):

$$[T] = T^{1},\quad [G]=L^{3}M^{-1}T^{-2},\quad [m]=M^{1},\quad [a]=L^{1}$$

Equating powers of the fundamental dimensions,

$$L: \; 0 = 3p + r$$ $$M: \; 0 = -p + q$$ $$T: \; 1 = -2p$$

Solving,

$$p = -\frac12,\quad q = -\frac12,\quad r = \frac32$$

Hence

$$T \propto G^{-1/2}\,m^{-1/2}\,a^{3/2} \;=\;k\,\sqrt{\dfrac{a^{3}}{G\,m}}$$

for some dimensionless constant $$k$$ that is the same for both experiments because the geometry is identical.

Initial situation: side $$=a$$, mass $$=m$$, time $$=T = 4\,$$s.

New situation: side $$=2a$$, mass $$=2m$$. Using the proportionality,

$$\dfrac{T'}{T}=\dfrac{\sqrt{(2a)^{3}/(G\,2m)}}{\sqrt{a^{3}/(G\,m)}}$$

$$\qquad=\,\dfrac{2^{3/2}}{2^{1/2}} = 2$$

Therefore $$T' = 2T = 2 \times 4\text{ s} = 8\text{ s}$$.

So the three spheres will collide after 8 seconds.

Solution :

For a satellite revolving around the earth in circular orbit :

$$K = \frac{GMm}{2r}$$

Given :

$$m = 1000\text{ kg}$$

$$M = 6 \times 10^{24}\text{ kg}$$

$$G = 6.67 \times 10^{-11}\text{ Nm}^2\text{kg}^{-2}$$

Radius of earth :

$$R = 6.4 \times 10^6\text{ m}$$

Height of satellite :

$$h = 270\text{ km}$$

$$= 2.7 \times 10^5\text{ m}$$

Orbital radius :

$$r = R+h$$

$$= 6.4 \times 10^6 + 2.7 \times 10^5$$

$$= 6.67 \times 10^6\text{ m}$$

Substituting values :

$$K = \frac{(6.67 \times 10^{-11})(6 \times 10^{24})(1000)}{2(6.67 \times 10^6)}$$

$$= \frac{6 \times 10^{16}}{2 \times 10^6}$$

$$= 3 \times 10^{10}\text{ J}$$

Final Answer :

$$3$$

If diameter reduced to 1/3, radius R/3, mass same. g' = GM/(R/3)² = 9GM/R² = 9g.

The answer is 9.

The gravitational potential energy due to the central mass $$M$$ is
$$V_g(r)= -\dfrac{G M m}{r}\,.$$

For a circular orbit of radius $$r_0$$ under gravity alone, the inward (centripetal) force balance is
$$\dfrac{m v_0^{\,2}}{r_0}= \dfrac{G M m}{r_0^{2}} \;\;\Longrightarrow\;\; v_0^{\,2}= \dfrac{G M}{r_0}\,.$$

The corresponding time period is therefore
$$T_0 =\dfrac{2\pi r_0}{v_0}=2\pi\sqrt{\dfrac{r_0^{3}}{G M}} \;\;\Longrightarrow\;\; T_0^{2}= \dfrac{4\pi^{2} r_0^{3}}{G M}\,.$$

Now an additional central potential is applied:
$$V_c(r)=\dfrac{m\alpha}{r^{3}}\qquad (\alpha\gt 0).$$

The radial force associated with this potential is obtained from $$F_r=-\dfrac{dV_c}{dr}$$:
$$F_c(r)=-\frac{d}{dr}\left(\dfrac{m\alpha}{r^{3}}\right)=+\,\dfrac{3m\alpha}{r^{4}}.$$
Since this force is positive in the outward radial direction, it is repulsive. Hence it opposes gravitation.

To keep the particle in the same circular orbit of radius $$r_0$$, the inward gravitational pull must now be reduced by this repulsion:
$$\dfrac{m v_1^{\,2}}{r_0}= \dfrac{G M m}{r_0^{2}}-\dfrac{3m\alpha}{r_0^{4}} \;\;\Longrightarrow\;\; v_1^{\,2}= \dfrac{G M}{r_0}-\dfrac{3\alpha}{r_0^{3}}.$$

The new time period becomes
$$T_1 =\dfrac{2\pi r_0}{v_1}\,,\qquad T_1^{2}= \dfrac{4\pi^{2} r_0^{2}}{v_1^{\,2}} =\dfrac{4\pi^{2} r_0^{3}} {\,G M-\dfrac{3\alpha}{r_0^{2}}}\;.$$ Define $$A=4\pi^{2}r_0^{3}$$ for brevity, so that
$$T_0^{2}= \dfrac{A}{G M},\qquad T_1^{2}= \dfrac{A}{\,G M-\dfrac{3\alpha}{r_0^{2}}}\;.$$

We now evaluate the required ratio:
$$\dfrac{T_1^{2}-T_0^{2}}{T_1^{2}} =1-\dfrac{T_0^{2}}{T_1^{2}} =1-\dfrac{\dfrac{A}{G M}} {\dfrac{A}{\,G M-\dfrac{3\alpha}{r_0^{2}}}} =1-\dfrac{G M-\dfrac{3\alpha}{r_0^{2}}}{G M} =\dfrac{3\alpha}{G M r_0^{2}}\;.$$

Thus
$$\boxed{\dfrac{T_1^{2}-T_0^{2}}{T_1^{2}}=\dfrac{3\alpha}{G M r_0^{2}}}\,.$$
Hence the correct option is:
Option A which is: $$\dfrac{3\alpha}{G M r_0^{2}}.$$

Satellite mass = $$10^3$$ kg, initial orbit radius = $$2R$$.

Total energy in circular orbit: $$E = -\frac{GMm}{2r}$$.

Initial energy: $$E_1 = -\frac{GM \times 10^3}{2 \times 2R} = -\frac{GM \times 10^3}{4R}$$.

Using $$GM = gR^2$$: $$E_1 = -\frac{gR \times 10^3}{4} = -\frac{10 \times R \times 10^3}{4} = -2500R$$.

Energy supplied: $$\frac{10^4R}{6}$$ J.

New energy: $$E_2 = E_1 + \frac{10^4R}{6} = -2500R + \frac{10000R}{6} = R(-2500 + \frac{5000}{3}) = R\frac{-7500+5000}{3} = \frac{-2500R}{3}$$.

$$E_2 = -\frac{GMm}{2r_2} = -\frac{gR^2 \times 10^3}{2r_2} = -\frac{10^4R^2}{2r_2}$$.

$$\frac{-2500R}{3} = \frac{-10^4R^2}{2r_2}$$

$$r_2 = \frac{10^4R^2 \times 3}{2 \times 2500R} = \frac{3 \times 10^4R}{5000} = 6R$$.

The correct answer is Option 4: $$6R$$.

A 90 kg body is placed at a height $$2R$$ above Earth’s surface and the gravitational pull on it is to be determined.

The acceleration due to gravity at a height $$h$$ above Earth’s surface is given by $$ g' = g\left(\frac{R}{R + h}\right)^2 $$. This follows from Newton’s law of gravitation, since $$g' = \frac{GM}{(R+h)^2}$$ and $$g = \frac{GM}{R^2}$$.

For $$h = 2R$$, the distance from Earth’s center is $$R + h = 3R$$, so substituting into the formula yields $$ g' = g\left(\frac{R}{3R}\right)^2 = g\times\frac{1}{9} = \frac{10}{9}\,\text{m/s}^2 $$.

The gravitational force on the body is then $$ F = m\,g' = 90 \times \frac{10}{9} = 100\,\text{N} $$.

The correct answer is Option A: 100 N.

We need to find the new orbital period when the distance from the Sun is reduced to one-fourth.

According to Kepler’s Third Law, $$T^2 \propto r^3$$, where $$T$$ is the orbital period and $$r$$ is the orbital radius.

Applying this to the initial and new conditions, we get $$\frac{T_2^2}{T_1^2} = \frac{r_2^3}{r_1^3} = \left(\frac{r_2}{r_1}\right)^3 = \left(\frac{1}{4}\right)^3 = \frac{1}{64}$$, which implies $$\frac{T_2}{T_1} = \frac{1}{8}$$.

Since the original period $$T_1$$ is 200 days, the new period is $$T_2 = \frac{T_1}{8} = \frac{200}{8} = 25$$ days.

The correct answer is Option (1): 25 days.

We need to find the weight of a body at a depth of $$R/4$$ below the Earth's surface, given that it weighs 300 N on the surface.

We know that the acceleration due to gravity at a depth $$d$$ below the Earth's surface (assuming uniform density) is $$g' = g\left(1 - \frac{d}{R}\right)$$, where $$g$$ is the acceleration due to gravity at the surface and $$R$$ is the radius of the Earth. This formula arises because at depth $$d$$, only the spherical shell of radius $$(R - d)$$ contributes to the gravitational field (by the shell theorem, the outer shell exerts no net force).

Substituting $$d = R/4$$ into this expression gives $$g' = g\left(1 - \frac{R/4}{R}\right) = g\left(1 - \frac{1}{4}\right) = g \times \frac{3}{4}$$.

Therefore, the weight of the body at this depth is $$W' = m \times g' = m \times \frac{3g}{4}$$.

Since the surface weight is $$W = mg = 300 \,\text{N}$$, it follows that $$W' = \frac{3}{4} \times W = \frac{3}{4} \times 300 = 225 \,\text{N}$$.

The correct answer is Option (4): 225 N.

At height $$h$$ above surface: $$g_h = g\frac{R^2}{(R+h)^2}$$.

At depth $$d$$ below surface: $$g_d = g\left(1 - \frac{d}{R}\right)$$.

Setting equal and using $$h = d$$:

$$\frac{R^2}{(R+h)^2} = 1 - \frac{h}{R} = \frac{R-h}{R}$$

$$R^3 = (R-h)(R+h)^2$$

Let $$x = h/R$$: $$1 = (1-x)(1+x)^2 = (1-x^2)(1+x) = 1+x-x^2-x^3$$

$$x + x^2(-1) - x^3 = 0$$... Actually: $$x - x^2 - x^3 = 0$$, $$x(1-x-x^2) = 0$$.

$$x^2 + x - 1 = 0$$, $$x = \frac{-1+\sqrt{5}}{2}$$.

$$h = \frac{(\sqrt{5}-1)R}{2} = \frac{\sqrt{5}R - R}{2}$$.

The answer corresponds to Option (4).

Escape velocity: $$v_e = \sqrt{\frac{2GM}{R}}$$.

Given: $$R_p = \frac{R_e}{3}$$ and $$M_p = \frac{M_e}{6}$$.

$$\frac{v_p}{v_e} = \sqrt{\frac{M_p/R_p}{M_e/R_e}} = \sqrt{\frac{M_p \cdot R_e}{M_e \cdot R_p}} = \sqrt{\frac{(M_e/6) \cdot R_e}{M_e \cdot (R_e/3)}} = \sqrt{\frac{1/6}{1/3}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$$

$$v_p = \frac{11.2}{\sqrt{2}} = \frac{11.2 \times \sqrt{2}}{2} = 5.6\sqrt{2} \approx 7.92 \text{ km/s}$$

The answer is Option (4): $$\boxed{7.9 \text{ km s}^{-1}}$$.

The acceleration due to gravity on the surface of Earth is:

$$g = \frac{GM}{R^2}$$

If the diameter is halved, the radius becomes $$R' = R/2$$.

$$g' = \frac{GM}{(R/2)^2} = \frac{GM}{R^2/4} = \frac{4GM}{R^2} = 4g$$

The answer is $$4g$$, which corresponds to Option (4).

Given: Gravitational potential $$V = -5.12 \times 10^7 \text{ J kg}^{-1}$$, acceleration due to gravity $$g = 6.4 \text{ m s}^{-2}$$, and mean radius of Earth $$R = 6400 \text{ km}$$.

Key formulas:

At distance $$r$$ from the centre of Earth:

$$V = -\frac{GM}{r}$$ and $$g = \frac{GM}{r^2}$$

Find $$r$$ from the ratio $$V/g$$: $$\frac{|V|}{g} = \frac{GM/r}{GM/r^2} = r$$

$$r = \frac{5.12 \times 10^7}{6.4} = 8 \times 10^6 \text{ m} = 8000 \text{ km}$$

Find the height above the surface: $$h = r - R = 8000 - 6400 = 1600 \text{ km}$$

The correct answer is Option (1): $$1600 \text{ km}$$.

A wire of length L and mass M is bent into a semicircular arc. The radius of the semicircle is determined by $$\pi R = L \Rightarrow R = \frac{L}{\pi}$$.

The linear mass density is $$\lambda = \frac{M}{L}$$. A particle of mass m is placed at the centre, and by symmetry the net force is directed toward the midpoint of the arc along the axis of symmetry.

Consider an element $$d\theta$$ at angle $$\theta$$ from the axis of symmetry. Its mass is $$dm = \lambda R\,d\theta = \frac{M}{L} \cdot \frac{L}{\pi}\,d\theta = \frac{M}{\pi}\,d\theta$$. The gravitational force from this element is $$dF = \frac{Gm\,dm}{R^2}$$, and the component along the axis of symmetry is $$dF\cos\theta$$.

Integrating from $$-\pi/2$$ to $$\pi/2$$ gives the total force:

$$ F = \int_{-\pi/2}^{\pi/2} \frac{Gm}{R^2} \cdot \frac{M}{\pi}\cos\theta\,d\theta = \frac{GmM}{\pi R^2}[\sin\theta]_{-\pi/2}^{\pi/2} = \frac{GmM}{\pi R^2}(2) = \frac{2GmM}{\pi R^2} $$

Substituting $$R = L/\pi$$:

$$ F = \frac{2GmM}{\pi(L/\pi)^2} = \frac{2GmM}{\pi \cdot L^2/\pi^2} = \frac{2GmM\pi}{L^2} $$

The correct answer is Option 4: $$\frac{2GmM\pi}{L^2}$$.

A simple pendulum at height $$R$$ above Earth's surface has time period $$T_1 = 4$$ s. We need $$T_2$$ at height $$2R$$.

The time period of a simple pendulum is $$T = 2\pi\sqrt{\frac{l}{g'}}$$ where $$g'$$ is the effective gravitational acceleration at height $$h$$:

$$ g' = \frac{g}{(1 + h/R)^2} $$

At height $$h = R$$:

$$ g_1 = \frac{g}{(1+1)^2} = \frac{g}{4} $$

$$ T_1 = 2\pi\sqrt{\frac{l}{g/4}} = 2\pi\sqrt{\frac{4l}{g}} $$

At height $$h = 2R$$:

$$ g_2 = \frac{g}{(1+2)^2} = \frac{g}{9} $$

$$ T_2 = 2\pi\sqrt{\frac{l}{g/9}} = 2\pi\sqrt{\frac{9l}{g}} $$

Dividing to find the relation between $$T_2$$ and $$T_1$$:

$$ \frac{T_2}{T_1} = \frac{\sqrt{9}}{\sqrt{4}} = \frac{3}{2} $$

Therefore,

$$ T_2 = \frac{3}{2}T_1 $$

Multiplying both sides by 2:

$$ 2T_2 = 3T_1 $$

This corresponds to $$3T_1 = 2T_2$$.

The correct answer is Option (4): $$3T_1 = 2T_2$$.

Find the length of a seconds pendulum at height $$h = 2R$$ where $$g = \pi^2$$ m/s$$^2$$.

At height $$h=2R$$ above Earth’s surface the acceleration due to gravity becomes $$g' = g\left(\frac{R}{R+h}\right)^2 = \pi^2 \left(\frac{R}{R+2R}\right)^2 = \pi^2 \times \frac{1}{9} = \frac{\pi^2}{9}$$ m/s$$^2$$.

A seconds pendulum has a time period $$T = 2$$ s and its period satisfies $$T = 2\pi\sqrt{\frac{l}{g'}}$$. Substituting $$T=2$$ and $$g'=\pi^2/9$$ gives $$2 = 2\pi\sqrt{\frac{l}{\pi^2/9}}$$. Inside the square root one finds $$\sqrt{\frac{l}{\pi^2/9}} = \sqrt{\frac{9l}{\pi^2}} = \frac{3\sqrt{l}}{\pi}$$. Hence $$2 = 2\pi\times\frac{3\sqrt{l}}{\pi} = 6\sqrt{l}$$, which leads to $$6\sqrt{l} = 2$$, so $$\sqrt{l} = \frac{2}{6} = \frac{1}{3}$$ and therefore $$l = \frac{1}{9}$$ m.

The correct answer is Option B: $$\frac{1}{9}$$ m.

Four identical particles of mass $$m$$ are at the corners of a square of side $$a$$. The net gravitational force on one particle due to the other three is given as $$\frac{(2\sqrt{2}+1)}{32}\frac{Gm^2}{L^2}$$. We need to find $$a$$ in terms of $$L$$.

Consider the particle at one corner. The other three particles are two adjacent particles at distance $$a$$ (along the sides) and one diagonal particle at distance $$a\sqrt{2}$$. Each adjacent particle exerts a force: $$F_{adj} = \frac{Gm^2}{a^2}$$. The two adjacent forces are perpendicular to each other along the two sides of the square, so by the Pythagorean theorem the resultant is $$F_{resultant,adj} = \sqrt{F_{adj}^2 + F_{adj}^2} = F_{adj}\sqrt{2} = \frac{\sqrt{2}\cdot Gm^2}{a^2}$$. This resultant points along the diagonal toward the diagonally opposite corner.

The diagonal particle exerts a force $$F_{diag} = \frac{Gm^2}{(a\sqrt{2})^2} = \frac{Gm^2}{2a^2}$$, which also points along the same diagonal direction. Since all forces are along the diagonal, the total net force is $$F_{net} = \frac{\sqrt{2}\cdot Gm^2}{a^2} + \frac{Gm^2}{2a^2} = \frac{Gm^2}{a^2}\left(\sqrt{2} + \tfrac{1}{2}\right) = \frac{(2\sqrt{2}+1)}{2}\cdot\frac{Gm^2}{a^2}$$. Equating this to the given expression $$\frac{(2\sqrt{2}+1)}{2}\cdot\frac{Gm^2}{a^2} = \frac{(2\sqrt{2}+1)}{32}\cdot\frac{Gm^2}{L^2}$$, the common factor $$(2\sqrt{2}+1)\,Gm^2$$ cancels, giving $$\frac{1}{2a^2} = \frac{1}{32L^2}$$, and hence $$2a^2 = 32L^2$$, $$a^2 = 16L^2$$, $$a = 4L$$.

The correct answer is Option (2): $$4L$$.

For a uniform solid sphere of mass M and radius R:

Gravitational potential at the surface:

$$V_{surface} = -\frac{GM}{R}$$

Gravitational potential at the centre:

$$V_{centre} = -\frac{3GM}{2R}$$

Since $$V = V_{surface} = -\frac{GM}{R}$$:

$$V_{centre} = -\frac{3GM}{2R} = \frac{3}{2}\left(-\frac{GM}{R}\right) = \frac{3}{2}V$$

The gravitational potential at the centre is $$\frac{3V}{2}$$.

The acceleration due to gravity at a distance $$r$$ from the centre of the Earth is

$$g(r)=\frac{GM}{r^{2}}$$

where $$G$$ is the gravitational constant and $$M$$ is the mass of the Earth.

For a satellite at height $$h$$ above the Earth’s surface (Earth’s radius $$R$$), the distance from the centre of the Earth is

$$r = R + h$$

Hence, for satellites P and Q we have

$$g_P = \frac{GM}{(R+h_P)^{2}}, \qquad g_Q = \frac{GM}{(R+h_Q)^{2}}$$

Taking their ratio, the common factor $$GM$$ cancels:

$$\frac{g_P}{g_Q} = \frac{(R+h_Q)^{2}}{(R+h_P)^{2}}$$ $$-(1)$$

Given data:

• $$h_P = \frac{R}{3} \;\;\Rightarrow\;\; R+h_P = R + \frac{R}{3} = \frac{4R}{3}$$
• $$\dfrac{g_P}{g_Q} = \dfrac{36}{25}$$

Substitute in equation $$(1)$$:

$$\frac{36}{25} = \frac{(R+h_Q)^{2}}{\left(\dfrac{4R}{3}\right)^{2}}$$

Solve for $$R+h_Q$$:

$$R+h_Q = \frac{6}{5}\left(\frac{4R}{3}\right) = \frac{24R}{15} = \frac{8R}{5}$$

Therefore,

$$h_Q = \frac{8R}{5} - R = \frac{8R}{5} - \frac{5R}{5} = \frac{3R}{5}$$

Hence, the height of satellite Q above the Earth’s surface is $$\displaystyle \frac{3R}{5}$$.

Option A which is: $$3R/5$$

We have an object falling from height $$R$$ above the Earth's surface (where $$R$$ is the radius of Earth). We use conservation of energy to find the velocity at the surface.

The gravitational potential energy at a distance $$r$$ from the center of Earth is $$U = -\frac{GMm}{r}$$. At the initial position (distance $$2R$$ from center), the object is at rest, so

$$E_i = -\frac{GMm}{2R}$$

At the Earth's surface (distance $$R$$ from center):

$$E_f = \frac{1}{2}mv^2 - \frac{GMm}{R}$$

By conservation of energy, $$E_i = E_f$$, so

$$-\frac{GMm}{2R} = \frac{1}{2}mv^2 - \frac{GMm}{R}$$

Rearranging:

$$\frac{1}{2}mv^2 = \frac{GMm}{R} - \frac{GMm}{2R} = \frac{GMm}{2R}$$

This gives $$v^2 = \frac{GM}{R}$$. Now since $$g = \frac{GM}{R^2}$$, we have $$GM = gR^2$$, so

$$v^2 = \frac{gR^2}{R} = gR$$

So, the answer is $$v = \sqrt{gR}$$.

Statement I: True. Earth's rotation affects g: $$g' = g - \omega^2 R\cos^2\lambda$$.

Statement II: The effect is maximum at the equator ($$\cos^2 0° = 1$$) and minimum (zero) at the poles. So Statement II says the opposite, which is false.

The correct answer is Option 4: Statement I is true but Statement II is false.

The escape velocities of planets A and B are in ratio 1:2, and their radii are in ratio 1:3. We need the ratio of their gravitational accelerations.

To begin,

$$ v_e = \sqrt{2gR} $$

Squaring: $$v_e^2 = 2gR$$, which gives $$g = \frac{v_e^2}{2R}$$.

Next,

$$ \frac{v_{eA}^2}{v_{eB}^2} = \frac{g_A R_A}{g_B R_B} $$

Given $$\frac{v_{eA}}{v_{eB}} = \frac{1}{2}$$ and $$\frac{R_A}{R_B} = \frac{1}{3}$$:

$$ \left(\frac{1}{2}\right)^2 = \frac{g_A}{g_B} \times \frac{1}{3} $$

$$ \frac{1}{4} = \frac{g_A}{3g_B} $$

$$ \frac{g_A}{g_B} = \frac{3}{4} $$

The ratio of acceleration due to gravity is $$\dfrac{3}{4}$$.

The correct answer is Option 4: $$\dfrac{3}{4}$$.

A body is released from height $$h = R$$ above the Earth's surface. Using conservation of energy:

$$-\frac{GMm}{R+R} = \frac{1}{2}mv^2 - \frac{GMm}{R}$$

$$\frac{1}{2}mv^2 = \frac{GMm}{R} - \frac{GMm}{2R} = \frac{GMm}{2R}$$

$$v^2 = \frac{GM}{R}$$

Since $$g = \frac{GM}{R^2}$$, we have $$GM = gR^2$$:

$$v^2 = \frac{gR^2}{R} = gR$$

$$v = \sqrt{gR}$$

This matches option 2.

Potential energy at a distance r from the centre of earth is given by :

$$U = -\frac{GMm}{r}$$

Initial potential energy at earth surface :

$$U_1 = -\frac{GMm}{R_e}$$

Given height,

$$h = 2R_e$$

Therefore, final distance from centre of earth :

$$r = R_e + 2R_e = 3R_e$$

Final potential energy :

$$U_2 = -\frac{GMm}{3R_e}$$

Increase in potential energy :

$$\Delta U = U_2 - U_1$$

$$= -\frac{GMm}{3R_e} + \frac{GMm}{R_e}$$

$$= \frac{2GMm}{3R_e}$$

Using,

$$g = \frac{GM}{R_e^2}$$

⇒

$$GM = gR_e^2$$

Substituting,

$$\Delta U = \frac{2mgR_e^2}{3R_e}$$

$$\Delta U = \frac{2}{3}mgR_e$$

Final Answer :

$$\frac{2}{3}mgR_e$$

A body of weight $$W$$ is at a height equal to 9 times the radius of earth ($$h = 9R$$).

Gravitational acceleration at height $$h$$.

Weight at that height.

The answer is Option B: $$\frac{W}{100}$$.

A spaceship in a circular orbit close to Earth's surface needs additional velocity to escape. We need to find this additional velocity.

First, calculate the orbital velocity for a circular orbit close to Earth's surface.

For a circular orbit at radius $$R$$ (Earth's radius, since the orbit is close to the surface), the gravitational force provides the centripetal force:

$$ \frac{mv_o^2}{R} = \frac{mg}{1} \implies v_o = \sqrt{gR} $$

Substituting $$g = 10$$ m/s$$^2$$ and $$R = 6400$$ km = $$6.4 \times 10^6$$ m:

$$ v_o = \sqrt{10 \times 6.4 \times 10^6} = \sqrt{64 \times 10^6} = 8000\;\text{m/s} = 8\;\text{km/s} $$

Next, calculate the escape velocity from this orbit.

The escape velocity from the surface of Earth is:

$$ v_e = \sqrt{2gR} = \sqrt{2} \times \sqrt{gR} = \sqrt{2} \times v_o $$

$$ v_e = 8\sqrt{2}\;\text{km/s} $$

Now, find the additional velocity needed.

The spaceship is already moving at orbital velocity $$v_o$$. To escape, it needs to reach escape velocity $$v_e$$. The additional velocity required is:

$$ \Delta v = v_e - v_o = 8\sqrt{2} - 8 = 8(\sqrt{2} - 1)\;\text{km/s} $$

Note: The mass of the spaceship ($$2 \times 10^4$$ kg) is not needed for this calculation, as the velocities are independent of mass.

The correct answer is Option 2: $$8(\sqrt{2} - 1)$$ km s$$^{-1}$$.

We need to find the depth $$d$$ below the Earth's surface where the gravitational acceleration equals four times its value at height $$3R$$ above the surface.

To begin,

At a height $$h$$ above the Earth's surface, the acceleration due to gravity is:

$$ g_h = \frac{g}{\left(1 + \frac{h}{R}\right)^2} $$

This follows from Newton's law of gravitation: $$g_h = \frac{GM}{(R+h)^2} = \frac{g R^2}{(R+h)^2}$$, where $$g = GM/R^2$$ is the surface value.

Next,

$$ g_h = \frac{g}{\left(1 + \frac{3R}{R}\right)^2} = \frac{g}{(1+3)^2} = \frac{g}{16} $$

From this,

At depth $$d$$, assuming uniform density of the Earth:

$$ g_d = g\left(1 - \frac{d}{R}\right) $$

This is because only the mass in the sphere of radius $$(R-d)$$ contributes to gravity at that depth (by the shell theorem), and that mass is proportional to $$(R-d)^3$$ while the distance from the centre is $$(R-d)$$, giving $$g_d = g(R-d)/R$$.

Continuing,

$$ g\left(1 - \frac{d}{R}\right) = 4 \times \frac{g}{16} $$

$$ g\left(1 - \frac{d}{R}\right) = \frac{g}{4} $$

Dividing both sides by $$g$$:

$$ 1 - \frac{d}{R} = \frac{1}{4} $$

$$ \frac{d}{R} = 1 - \frac{1}{4} = \frac{3}{4} $$

Now,

$$ d = \frac{3}{4} \times R = \frac{3}{4} \times 6400 = 4800\;\text{km} $$

The depth is 4800 km.

The correct answer is Option 4: 4800 km.

Statement I: Acceleration due to gravity decreases as you go 'up' or 'down' from earth's surface.

Above the surface: $$g_h = g\left(\frac{R}{R+h}\right)^2$$ which decreases with height $$h$$.

Below the surface: $$g_d = g\left(1 - \frac{d}{R}\right)$$ which decreases with depth $$d$$.

So Statement I is correct.

Statement II: Acceleration due to gravity is same at height $$h$$ and depth $$d$$ if $$h = d$$.

At height $$h$$: $$g_h = g\left(1 - \frac{2h}{R}\right)$$ (for small $$h$$)

At depth $$d$$: $$g_d = g\left(1 - \frac{d}{R}\right)$$

For $$g_h = g_d$$: $$1 - \frac{2h}{R} = 1 - \frac{d}{R}$$, which gives $$d = 2h$$.

So $$g$$ is the same when $$d = 2h$$, not when $$h = d$$. Statement II is incorrect.

The correct answer is Option 3: Statement I is correct but Statement II is incorrect.

We are given that Earth has mass $$M_e = 9M_p$$ and radius $$R_e = 2R_p$$, where $$M_p$$ and $$R_p$$ are the mass and radius of planet $$P$$.

The escape velocity on Earth is $$v_e = \sqrt{\frac{2GM_e}{R_e}}$$, and on planet $$P$$ it is $$v_p = \sqrt{\frac{2GM_p}{R_p}}$$.

From the given relations, $$M_p = \frac{M_e}{9}$$ and $$R_p = \frac{R_e}{2}$$. Substituting these in:

$$v_p = \sqrt{\frac{2G \cdot \frac{M_e}{9}}{\frac{R_e}{2}}} = \sqrt{\frac{2GM_e \cdot 2}{9R_e}} = \sqrt{\frac{2}{9}} \cdot \sqrt{\frac{2GM_e}{R_e}}$$

So $$v_p = \frac{\sqrt{2}}{3} \cdot v_e = \frac{v_e}{3}\sqrt{2}$$.

Comparing with the given form $$\frac{v_e}{3}\sqrt{x}$$, we get $$x = 2$$. Hence, the correct answer is $$2$$.

For a satellite in circular orbit, the orbital angular momentum is $$L = mvr$$, where $$v$$ is the orbital velocity and $$r$$ is the distance from the Earth's centre.

For a circular orbit, gravitational force provides centripetal force, so

$$\frac{GMm}{r^2} = \frac{mv^2}{r}$$

which gives $$v = \sqrt{\frac{GM}{r}}$$.

Substituting into the angular momentum expression:

$$L = mr\sqrt{\frac{GM}{r}} = m\sqrt{GMr}$$

So $$L \propto \sqrt{r}$$. Now, the problem says the distance is increased by eight times to its initial value, meaning the new distance is $$r' = r + 8r = 9r$$ (increased by eight times the original).

The new angular momentum is then

$$\frac{L'}{L} = \sqrt{\frac{9r}{r}} = \sqrt{9} = 3$$

So $$L' = 3L$$. Hence, the correct answer is Option 4.

The acceleration due to gravity at the surface of a planet is given by:

$$g = \frac{GM}{R^2}$$

Since $$M = \frac{4}{3}\pi R^3 \rho$$, we get:

$$g = \frac{G \times \frac{4}{3}\pi R^3 \rho}{R^2} = \frac{4}{3}\pi G R \rho$$

So $$g \propto R\rho$$.

For Planet A: Radius = $$R$$, Density = $$\rho$$

$$g_A = \frac{4}{3}\pi G R \rho$$

For Planet B: Radius = $$4R$$, Density = $$\frac{\rho}{3}$$

$$g_B = \frac{4}{3}\pi G (4R) \left(\frac{\rho}{3}\right) = \frac{4}{3}\pi G \times \frac{4R\rho}{3}$$

The ratio:

$$\frac{g_A}{g_B} = \frac{R\rho}{\frac{4R\rho}{3}} = \frac{R\rho \times 3}{4R\rho} = \frac{3}{4}$$

Therefore, $$g_A : g_B = 3 : 4$$.

We need to find the time period of a satellite revolving at a height $$h = R$$ above the Earth's surface, where $$R$$ is the radius of the Earth and $$g = \pi^2$$ m/s$$^2$$. The satellite is at height $$h = R$$ above the Earth's surface so the orbital radius (distance from the center of the Earth) is
$$r = R + h = R + R = 2R$$.

For a satellite of mass $$m$$ in a circular orbit the gravitational force is balanced by the centripetal force:
$$\frac{GMm}{r^2} = \frac{mv^2}{r}$$ which gives the orbital velocity $$v = \sqrt{\frac{GM}{r}}$$. The time period is the circumference divided by velocity:
$$T = \frac{2\pi r}{v} = 2\pi r \cdot \sqrt{\frac{r}{GM}} = 2\pi\sqrt{\frac{r^3}{GM}}$$. At the Earth's surface the gravitational acceleration satisfies $$g = \frac{GM}{R^2}$$, which gives $$GM = gR^2$$. Substituting this into the time period formula yields
$$T = 2\pi\sqrt{\frac{r^3}{gR^2}}$$.

Substituting $$r = 2R$$ gives
$$T = 2\pi\sqrt{\frac{(2R)^3}{gR^2}} = 2\pi\sqrt{\frac{8R^3}{gR^2}} = 2\pi\sqrt{\frac{8R}{g}}$$. With $$g = \pi^2$$ we get
$$T = 2\pi\sqrt{\frac{8R}{\pi^2}} = 2\pi \cdot \frac{\sqrt{8R}}{\pi} = 2\sqrt{8R}$$. Since $$\sqrt{8} = 2\sqrt{2}$$, it follows that
$$T = 2 \cdot 2\sqrt{2R} = 4\sqrt{2R}$$ which can also be written as
$$T = \sqrt{16 \cdot 2R} = \sqrt{32R}$$.

The correct answer is Option 3: $$\sqrt{32R}$$.

We need the weight at an altitude of 3200 km given that the weight on the surface is 18 N and $$R_e = 6400$$ km.

From the relation between gravitational acceleration at altitude h and at the surface, namely $$g_h = \frac{GM}{(R_e+h)^2}$$ and $$g = \frac{GM}{R_e^2}$$, it follows that the weight at altitude h is given by the formula

$$ W' = W\left(\frac{R_e}{R_e + h}\right)^2 $$

Substituting the values into this expression, we obtain

$$ W' = 18 \times \left(\frac{6400}{6400 + 3200}\right)^2 = 18 \times \left(\frac{6400}{9600}\right)^2 = 18 \times \left(\frac{2}{3}\right)^2 = 18 \times \frac{4}{9} = 8\;\text{N} $$

Hence, the weight at an altitude of 3200 km is 8 N, so the correct answer is Option 4: 8 N.

The weight of a body on the surface of the earth is $$100$$ N. We need to find the gravitational force at height $$h = \dfrac{R}{4}$$ from the surface.

$$g' = g \cdot \dfrac{R^2}{(R+h)^2}$$

$$g' = g \cdot \dfrac{R^2}{(R + R/4)^2} = g \cdot \dfrac{R^2}{(5R/4)^2} = g \cdot \dfrac{R^2}{25R^2/16} = g \cdot \dfrac{16}{25}$$

$$W' = mg' = mg \cdot \dfrac{16}{25} = 100 \times \dfrac{16}{25} = 64 \text{ N}$$

The gravitational force at height $$R/4$$ is $$64$$ N, which corresponds to Option A.

Two particles of equal mass $$m$$ move in a circle of radius $$r$$ under their mutual gravitational attraction. They are diametrically opposite on the circle, so the distance between them is $$2r$$.

$$F = \frac{Gm^2}{(2r)^2} = \frac{Gm^2}{4r^2}$$

This gravitational force provides the centripetal force for each particle moving in a circle of radius $$r$$:

$$\frac{Gm^2}{4r^2} = \frac{mv^2}{r}$$

$$\frac{Gm}{4r^2} = \frac{v^2}{r}$$

$$v^2 = \frac{Gm}{4r}$$

$$v = \sqrt{\frac{Gm}{4r}}$$

The correct answer is Option D: $$\sqrt{\dfrac{Gm}{4r}}$$.

Acceleration due to gravity at the surface of a planet: $$g = \frac{GM}{R^2} = \frac{G \cdot \frac{4}{3}\pi R^3 \rho}{R^2} = \frac{4}{3}G\pi R\rho$$

For planet A: $$g_A = \frac{4}{3}G\pi R\rho$$

For planet B: $$g_B = \frac{4}{3}G\pi (1.5R)\left(\frac{\rho}{2}\right) = \frac{4}{3}G\pi \times \frac{1.5R\rho}{2} = \frac{4}{3}G\pi \times \frac{3R\rho}{4}$$

$$\frac{g_B}{g_A} = \frac{3R\rho/4}{R\rho} = \frac{3}{4}$$

The ratio $$g_B : g_A = 3 : 4$$.

This matches option 3.

$$\frac{G M_E m}{r^2} = \frac{m v_0^2}{r} \implies v_0 = \sqrt{\frac{G M_E}{r}}$$

$$\text{Orbital speed } v_0 \text{ is independent of the satellite's mass } m \text{, so it remains identical for both } A \text{ and } B.$$

$$K = \frac{G M_E m}{2r}, \quad U = -\frac{G M_E m}{r}, \quad E = -\frac{G M_E m}{2r}$$

Since $$m_A = 2m_B$$, the kinetic, potential, and total energies of satellite A are twice those of satellite B

Two satellites of masses $$m$$ and $$3m$$ revolve in circular orbits of radii $$r$$ and $$3r$$ respectively. We need to find the ratio of their orbital speeds.

First, derive the orbital speed formula.

For a satellite in circular orbit, the gravitational force provides the centripetal force:

$$ \frac{GMm_s}{r_s^2} = \frac{m_s v^2}{r_s} $$

where $$M$$ is Earth's mass, $$m_s$$ is the satellite mass, and $$r_s$$ is the orbital radius. Notice that the satellite mass $$m_s$$ cancels:

$$ v = \sqrt{\frac{GM}{r_s}} $$

This is a crucial result: orbital speed depends only on the orbital radius, not on the satellite's mass.

Next, calculate the ratio of orbital speeds.

For satellite 1 (orbit radius $$r$$): $$v_1 = \sqrt{\frac{GM}{r}}$$

For satellite 2 (orbit radius $$3r$$): $$v_2 = \sqrt{\frac{GM}{3r}}$$

$$ \frac{v_1}{v_2} = \sqrt{\frac{GM/r}{GM/3r}} = \sqrt{\frac{3r}{r}} = \sqrt{3} $$

The ratio of orbital speeds is $$\sqrt{3} : 1$$.

The correct answer is Option 1: $$\sqrt{3} : 1$$.

A planet has mass $$M_p = 2M_e$$ and average density $$\rho_p = \rho_e$$. We need the weight of an object on this planet.

Since density is the same: $$\rho = \dfrac{M}{\frac{4}{3}\pi R^3}$$.

$$\dfrac{M_p}{\frac{4}{3}\pi R_p^3} = \dfrac{M_e}{\frac{4}{3}\pi R_e^3}$$.

$$\dfrac{R_p^3}{R_e^3} = \dfrac{M_p}{M_e} = 2 \implies R_p = 2^{1/3}\,R_e$$.

$$\dfrac{g_p}{g_e} = \dfrac{M_p}{M_e} \cdot \dfrac{R_e^2}{R_p^2} = 2 \cdot \dfrac{R_e^2}{2^{2/3}\,R_e^2} = 2 \cdot 2^{-2/3} = 2^{1/3}$$.

$$W_p = \dfrac{g_p}{g_e} \cdot W = 2^{1/3}\,W$$.

The correct answer is Option A: $$2^{1/3}W$$.

We need to find the weight of a body at depth $$d = R/2$$ below Earth's surface, given its surface weight is 200 N.

To begin,

Assuming Earth has uniform mass density, the acceleration due to gravity at depth $$d$$ below the surface is:

$$ g_d = g\left(1 - \frac{d}{R}\right) $$

This is because, by the shell theorem, only the spherical mass at radii less than $$(R - d)$$ contributes to gravity. This inner mass is proportional to $$(R-d)^3$$, and the gravitational acceleration at distance $$(R-d)$$ from the centre is $$g_d = g(R-d)/R$$.

Next,

$$ W' = W \times \frac{g_d}{g} = W\left(1 - \frac{d}{R}\right) = 200\left(1 - \frac{R/2}{R}\right) = 200\left(1 - \frac{1}{2}\right) = 200 \times \frac{1}{2} = 100\;\text{N} $$

The weight at depth $$R/2$$ is 100 N.

The correct answer is Option 2: 100 N.

We need to identify the incorrect statement from the given options.

Option A: The speed of satellite in a given circular orbit remains constant.

This is correct. In a circular orbit, the gravitational force is always perpendicular to the velocity, so it only changes the direction, not the magnitude of velocity. The speed remains constant.

Option B: For a planet revolving around the sun in an elliptical orbit, the total energy of the planet remains constant.

This is correct. The gravitational force is conservative, so the total mechanical energy (kinetic + potential) is conserved in any orbit.

Option C: The linear speed of a planet revolving around the sun remains constant.

This is INCORRECT. A planet moves in an elliptical orbit (Kepler's First Law). By Kepler's Second Law (equal areas in equal times), the planet moves faster at perihelion (closest to the sun) and slower at aphelion (farthest from the sun). The linear speed varies throughout the orbit.

Option D: When a body falls towards earth, the displacement of earth towards the body is negligible.

This is correct. By Newton's Third Law, the earth also accelerates toward the body, but since the earth's mass is enormous, its acceleration (and hence displacement) is negligibly small.

The incorrect statement is Option C.

According to Newton’s law of gravitation, the gravitational force between the sun and a planet is :

$$F = \frac{GMm}{r^2}$$

Hence, force is inversely proportional to square of distance from the sun.

Therefore,

A is correct.

The force is directly proportional to the product of masses of the sun and the planet, not inversely proportional.

Therefore,

B is incorrect.

Centripetal force acting on the planet is always directed towards the sun.

Therefore,

C is incorrect.

According to Kepler’s third law :

$$T^2 \propto a^3$$

where \(T\) is the time period and \(a\) is the semi-major axis of the elliptical orbit.

Therefore,

D is correct.

Final Answer :

A and D

The assertion states that a pendulum clock when taken to Mount Everest becomes fast.

It is reasoned that the value of g (acceleration due to gravity) is less at Mount Everest than on the surface of the earth.

Upon analysis, at higher altitude such as Mount Everest the gravitational acceleration decreases, so $$g$$ is indeed less there. Therefore, the reason is correct.

Analyzing the assertion further, we recall that the time period of a simple pendulum is:

$$T = 2\pi\sqrt{\frac{L}{g}}$$

Since $$g$$ decreases at Mount Everest, the time period $$T$$ increases (the pendulum oscillates more slowly). A longer time period means each swing takes more time, so the clock runs slow rather than fast, making the assertion incorrect.

The assertion is incorrect but the reason is correct, so the correct answer is Option D.

Consider Statement I, which says acceleration due to gravity is different at different places on the surface of Earth. This is true because the Earth is not a perfect sphere (it is an oblate spheroid), so the radius varies with latitude. Additionally, the rotational effect of the Earth reduces effective gravity at the equator compared to the poles. The effective gravity is given by $$g' = g - R\omega^2\cos^2\lambda$$, where $$\lambda$$ is the latitude. So Statement I is correct.

Now consider Statement II, which says acceleration due to gravity increases as we go down below the Earth's surface. Below the surface, the acceleration due to gravity is given by

$$g_d = g\left(1 - \frac{d}{R}\right)$$

where $$d$$ is the depth. As depth increases, $$g_d$$ actually decreases (reaching zero at the center of the Earth). So Statement II is false.

Hence, Statement I is true but Statement II is false.

We need to evaluate two statements about escape velocity from a planet.

Recall the formula for escape velocity.

The escape velocity from the surface of a planet is derived by equating the kinetic energy of the object to the gravitational potential energy at the surface:

$$ \frac{1}{2}mv_e^2 = \frac{GMm}{R} $$

where $$G$$ is the gravitational constant, $$M$$ is the mass of the planet, $$R$$ is its radius, and $$m$$ is the mass of the object. Solving for $$v_e$$:

$$ v_e = \sqrt{\frac{2GM}{R}} $$

This can also be written as $$v_e = \sqrt{2gR}$$, where $$g = GM/R^2$$ is the surface gravitational acceleration.

Now evaluate Statement I.

"If the ratio of mass of the planet to its radius ($$M/R$$) increases, the escape velocity increases."

From the formula $$v_e = \sqrt{2GM/R}$$, we can write:

$$ v_e = \sqrt{2G \cdot \frac{M}{R}} $$

Since $$G$$ is a constant, $$v_e$$ is directly proportional to $$\sqrt{M/R}$$. Therefore, if $$M/R$$ increases, $$v_e$$ will also increase. Statement I is correct.

Now evaluate Statement II.

"Escape velocity is independent of the radius of the planet."

From the formula $$v_e = \sqrt{2GM/R}$$, we can clearly see that $$v_e$$ depends on $$R$$ (the radius appears in the denominator under the square root). If we keep $$M$$ constant and change $$R$$, the escape velocity changes. For example, if $$R$$ is halved (planet compressed to smaller size with same mass), $$v_e$$ increases by a factor of $$\sqrt{2}$$.

Note: Escape velocity is independent of the mass of the escaping object ($$m$$ cancels out in the derivation), but it does depend on the radius of the planet. Statement II is incorrect.

The correct answer is Option 2: Statement I is correct but Statement II is incorrect.

We need to evaluate two statements about satellite energy.

Recall the energy relations for a satellite in orbit. For a satellite of mass $$m$$ orbiting Earth at radius $$r$$:

- Kinetic Energy: $$KE = \frac{GMm}{2r}$$

- Potential Energy: $$PE = -\frac{GMm}{r}$$

- Total Energy: $$E = KE + PE = -\frac{GMm}{2r}$$

Note that $$E$$ is negative.

Check Statement I — "PE = E/2": $$E/2 = -\frac{GMm}{4r}$$

But $$PE = -\frac{GMm}{r}$$

Since $$PE \neq E/2$$ (in fact, $$PE = 2E$$), Statement I is incorrect.

Check Statement II — "KE = |E|/2": $$|E| = \frac{GMm}{2r}$$

$$|E|/2 = \frac{GMm}{4r}$$

But $$KE = \frac{GMm}{2r} = |E|$$

Since $$KE = |E| \neq |E|/2$$, Statement II is incorrect.

The correct relationships are: $$PE = 2E$$ and $$KE = -E = |E|$$.

Since both statements are incorrect, the correct answer is Option D: Both Statement I and Statement II are incorrect.

We need to find the gravitational potential at $$r = 3$$ cm, given the gravitational field $$g = -\frac{K}{r^2}$$ and $$V(r=2) = 10$$ J/kg.

Relation between field and potential:

$$V(r) - V(r_0) = -\int_{r_0}^{r} g \, dr$$

Calculating the potential difference:

$$V(3) - V(2) = -\int_{2}^{3} \left(-\frac{K}{r^2}\right) dr = K\int_{2}^{3} \frac{dr}{r^2}$$

$$= K\left[-\frac{1}{r}\right]_{2}^{3} = K\left(-\frac{1}{3} + \frac{1}{2}\right) = K \times \frac{1}{6}$$

Substituting values:

$$K = 6$$ J cm/kg (note: distances are in cm)

$$V(3) = V(2) + \frac{K}{6} = 10 + \frac{6}{6} = 10 + 1 = 11$$ J/kg

The gravitational potential at $$r = 3$$ cm is $$11$$ J/kg.

The correct answer is Option 2: $$11$$.

We need to find the time period of a simple pendulum when taken to a height $$R$$ (equal to Earth's radius) above the Earth's surface.

Find the acceleration due to gravity at height $$h = R$$.

At height $$h$$ above the Earth's surface:

$$ g' = g \cdot \frac{R^2}{(R + h)^2} $$

At $$h = R$$:

$$ g' = g \cdot \frac{R^2}{(R + R)^2} = g \cdot \frac{R^2}{4R^2} = \frac{g}{4} $$

Find the new time period.

The time period of a simple pendulum is $$T = 2\pi\sqrt{\frac{L}{g}}$$.

At height $$R$$:

$$ T' = 2\pi\sqrt{\frac{L}{g'}} = 2\pi\sqrt{\frac{L}{g/4}} = 2\pi\sqrt{\frac{4L}{g}} = 2 \cdot 2\pi\sqrt{\frac{L}{g}} = 2T $$

Find the value of $$x$$.

Since $$T' = xT$$ and $$T' = 2T$$:

$$ x = 2 $$

The correct answer is Option B: $$2$$.

$$g' = \frac{GM}{(R+h)^2} = \frac{GM}{R^2\left(1 + \frac{h}{R}\right)^2} = g\left(1 + \frac{h}{R}\right)^{-2}$$

$$\text{Applying the Binomial Theorem for } \frac{h}{R} \ll 1: \left(1 + \frac{h}{R}\right)^{-2} \approx 1 - \frac{2h}{R}$$

$$g' = g\left(1 - \frac{2h}{R}\right)$$

Escape velocity: $$v_e = \sqrt{\frac{2GM}{R}}$$

$$\frac{v_p}{v_e} = \sqrt{\frac{M_p}{M_e} \times \frac{R_e}{R_p}} = \sqrt{\frac{16}{1} \times \frac{1}{4}} = \sqrt{4} = 2$$

Ratio = $$2 : 1$$

Solution :

Time period of satellite is related to orbital radius by Kepler’s third law :

$$T^2 \propto r^3$$

Therefore,

$$\left(\frac{T_2}{T_1}\right)^2 = \left(\frac{r_2}{r_1}\right)^3$$

Given :

$$T_1 = 24\text{ hours}$$

New separation :

$$r_2 = \frac{r_1}{4}$$

Substituting :

$$\left(\frac{T_2}{24}\right)^2 = \left(\frac{1}{4}\right)^3$$

$$\left(\frac{T_2}{24}\right)^2 = \frac{1}{64}$$

$$\frac{T_2}{24} = \frac{1}{8}$$

$$T_2 = \frac{24}{8}$$

$$= 3\text{ hours}$$

Final Answer :

$$3\text{ hours}$$

Two identical particles of mass $$m$$ go round a circle of radius $$a$$. They are diametrically opposite, so the distance between them is $$2a$$.

The gravitational force between them provides the centripetal force:

$$\frac{Gm^2}{(2a)^2} = m\omega^2 a$$

$$\frac{Gm}{4a^2} = \omega^2 a$$

$$\omega^2 = \frac{Gm}{4a^3}$$

$$\omega = \sqrt{\frac{Gm}{4a^3}}$$

This matches option 3.

Given: Planet has mass $$M_p = 9M_e$$ and radius $$R_p = 4R_e$$. Escape velocity on Earth $$V_e = 11.2 \times 10^3$$ m/s.

The escape velocity formula is:

$$V_{esc} = \sqrt{\frac{2GM}{R}}$$

The ratio of escape velocities:

$$\frac{V_p}{V_e} = \sqrt{\frac{M_p}{M_e} \cdot \frac{R_e}{R_p}} = \sqrt{\frac{9M_e}{M_e} \cdot \frac{R_e}{4R_e}} = \sqrt{\frac{9}{4}} = \frac{3}{2}$$

Therefore:

$$V_p = \frac{3}{2} \times V_e = \frac{3}{2} \times 11.2 = 16.8 \text{ km/s}$$

Therefore, the correct answer is Option B: $$\mathbf{16.8}$$.

Assertion A: Earth has atmosphere whereas moon doesn't have any atmosphere.

This is TRUE. Earth has a significant atmosphere while the moon has virtually no atmosphere.

Reason R: The escape velocity on moon is very small as compared to that on earth.

This is TRUE. The escape velocity on the moon is about $$2.38\;\text{km/s}$$, while on Earth it is about $$11.2\;\text{km/s}$$.

The escape velocity determines whether gas molecules can escape the gravitational pull of a celestial body. Since the moon's escape velocity is very low, gas molecules (which have thermal velocities comparable to or exceeding the moon's escape velocity) easily escape from the moon's surface. On Earth, the higher escape velocity retains the atmosphere.

Therefore, R is the correct explanation of A.

The correct answer is Both A and R are correct and R is the correct explanation of A.

Distance of Earth from Sun = $$1.5 \times 10^6$$ km and the period of the imaginary planet is 2.83 years.

According to Kepler’s Third Law, $$T^2 \propto R^3 \implies \left(\frac{T_2}{T_1}\right)^2 = \left(\frac{R_2}{R_1}\right)^3$$

Substituting Earth’s values $$T_1 = 1$$ year and $$R_1 = 1.5 \times 10^6$$ km into this relation gives $$\left(\frac{2.83}{1}\right)^2 = \left(\frac{R_2}{1.5 \times 10^6}\right)^3$$

Noting that $$2.83 \approx \sqrt{8}$$ so $$(2.83)^2 \approx 8$$ leads to $$8 = \left(\frac{R_2}{1.5 \times 10^6}\right)^3$$ and hence $$\frac{R_2}{1.5 \times 10^6} = \sqrt[3]{8} = 2$$

It follows that $$R_2 = 2 \times 1.5 \times 10^6 = 3 \times 10^6 \text{ km}$$

The correct answer is Option C: $$3 \times 10^6$$ km.

$$g' = g\left(1 - \frac{d}{R}\right) \implies W' = W\left(1 - \frac{d}{R}\right)$$

$$W' = 400\left(1 - \frac{R/2}{R}\right) = 400\left(1 - \frac{1}{2}\right) = 400\left(\frac{1}{2}\right) = 200\text{ N}$$

For a uniform solid sphere of mass $$M$$ and radius $$R$$ the escape speed from its surface is
$$v = \sqrt{\dfrac{2GM}{R}} \; \; -(1)$$

Initial data
Both stars have the same initial radius $$R$$.
Masses are related by $$M_B = 2M_A$$.
Since mass $$M = \dfrac{4}{3}\pi R^{3}\rho$$ for a sphere of density $$\rho$$, we have
$$\dfrac{\rho_B}{\rho_A} = \dfrac{M_B}{M_A} = 2 \;\;\; \Rightarrow \;\;\; \rho_B = 2\rho_A$$.

1. Final parameters of star A
Its radius is halved: $$R_A' = \dfrac{R}{2}$$.
Its density is still $$\rho_A$$, hence its final mass is
$$M_A' = \dfrac{4}{3}\pi \left(\dfrac{R}{2}\right)^3 \rho_A = \dfrac{1}{8}\left(\dfrac{4}{3}\pi R^{3}\rho_A\right)=\dfrac{M_A}{8}$$.
Mass lost by A
$$\Delta M = M_A - M_A' = M_A\left(1-\dfrac{1}{8}\right)=\dfrac{7M_A}{8}$$.

2. Final parameters of star B
The whole mass $$\Delta M$$ forms a spherical shell of density $$\rho_A$$ around B.
Let the outer radius after coating be $$R_B'$$.
Mass of the shell:
$$\dfrac{4}{3}\pi\bigl(R_B'^{3}-R^{3}\bigr)\rho_A = \dfrac{7M_A}{8} = \dfrac{7}{8}\left(\dfrac{4}{3}\pi R^{3}\rho_A\right).$$
Cancelling the common factor $$\dfrac{4}{3}\pi\rho_A$$ gives
$$R_B'^{3}-R^{3}= \dfrac{7}{8}R^{3}\;\;\;\Rightarrow\;\;\;R_B'^{3} =\left(1+\dfrac{7}{8}\right)R^{3}= \dfrac{15}{8}R^{3}.$$

Therefore
$$R_B' = R\left(\dfrac{15}{8}\right)^{1/3} \;\;\; -(2)$$
Final mass of B:
$$M_B' = M_B + \Delta M = 2M_A + \dfrac{7M_A}{8}= \dfrac{23M_A}{8} \;\;\; -(3)$$

3. Escape speeds after interaction
From (1):
Escape speed from A:
$$v_A = \sqrt{\dfrac{2G M_A'}{R_A'}} = \sqrt{\dfrac{2G\left(\dfrac{M_A}{8}\right)}{R/2}} = \sqrt{\dfrac{G M_A}{2R}} \;\;\; -(4)$$
Escape speed from B:
$$v_B = \sqrt{\dfrac{2G M_B'}{R_B'}} = \sqrt{\dfrac{2G\left(\dfrac{23M_A}{8}\right)} {R\left( \dfrac{15}{8}\right)^{1/3}}} \;\;\; -(5)$$

4. Required ratio
From (4) and (5):
$$\left(\dfrac{v_B}{v_A}\right)^2 = \dfrac{\,\dfrac{2G(23M_A/8)}{R(15/8)^{1/3}}\,}{\,\dfrac{G M_A}{2R}\,} = \left(\dfrac{2\cdot23}{8}\right)\cdot2\cdot\left(\dfrac{8}{15}\right)^{1/3} = \dfrac{23}{2}\left(\dfrac{8}{15}\right)^{1/3}.$$

Note that $$\left(\dfrac{8}{15}\right)^{1/3} = \dfrac{2^{3/3}}{15^{1/3}}= \dfrac{2}{15^{1/3}}.$$ Hence
$$\left(\dfrac{v_B}{v_A}\right)^2 = \dfrac{23}{2}\cdot\dfrac{2}{15^{1/3}} = \dfrac{23}{15^{1/3}}.$$

Taking the square root,
$$\dfrac{v_B}{v_A}= \dfrac{\sqrt{23}}{15^{1/6}} = \dfrac{\sqrt{10\,n}}{15^{1/6}},$$ which matches the given form $$\sqrt{\dfrac{10n}{15^{1/3}}}$$ because $$1/15^{1/6} = \sqrt{1/15^{1/3}}.$$

Equating the numerators:
$$\sqrt{10\,n} = \sqrt{23}\;\;\;\Rightarrow\;\;\;10n = 23 \;\;\;\Rightarrow\;\;\;n = 2.3.$$

The required value is 2.30.

We need to find $$\alpha$$ such that the acceleration due to gravity at height $$h$$ above the Earth's surface equals the acceleration due to gravity at depth $$\alpha h$$ below the surface, given $$h \ll R_E$$.

The acceleration due to gravity at a height $$h$$ above the surface (for $$h \ll R_E$$) is approximately $$g_h = g\left(1 - \dfrac{2h}{R_E}\right)$$.

The acceleration due to gravity at a depth $$d$$ below the surface is $$g_d = g\left(1 - \dfrac{d}{R_E}\right)$$.

Setting these equal with $$d = \alpha h$$: $$g\left(1 - \dfrac{2h}{R_E}\right) = g\left(1 - \dfrac{\alpha h}{R_E}\right)$$.

Cancelling $$g$$ and simplifying: $$-\dfrac{2h}{R_E} = -\dfrac{\alpha h}{R_E}$$, which gives $$\alpha = 2$$.

Hence, the correct answer is 2.

A body is projected vertically upwards with velocity $$v = \dfrac{v_e}{3}$$ (one-third of escape velocity), and we need to find the maximum height attained. The escape velocity from Earth's surface is $$v_e = \sqrt{2gR}$$ where $$R = 6400 \text{ km}$$ and $$g = 10 \text{ m s}^{-2}$$.

At the surface, the total energy equals kinetic plus potential energy: $$\dfrac{1}{2}mv^2 - \dfrac{GMm}{R} = -\dfrac{GMm}{R+h}$$.

Since $$v = \dfrac{v_e}{3} = \dfrac{\sqrt{2gR}}{3}$$, substituting into the energy equation yields $$\dfrac{1}{2}m \cdot \dfrac{2gR}{9} = \dfrac{GMm}{R} - \dfrac{GMm}{R+h}$$.

Using $$GM = gR^2$$, this gives $$\dfrac{mgR}{9} = mgR\left(1 - \dfrac{R}{R+h}\right) = mgR \cdot \dfrac{h}{R+h}$$. From this, $$\dfrac{1}{9} = \dfrac{h}{R+h}$$ so that $$R + h = 9h$$ and $$R = 8h$$, which leads to $$h = \dfrac{R}{8} = \dfrac{6400}{8} = 800 \text{ km}$$.

The correct answer is Option A: $$800 \text{ km}$$.

We know that the time period of a simple pendulum is $$T = 2\pi\sqrt{\frac{l}{g'}}$$, where $$g'$$ is the effective gravitational acceleration at that location. Since both clocks use identical pendulums (same length $$l$$), the ratio of their time periods depends only on the ratio of gravitational accelerations.

On the Earth's surface, $$g_{\text{earth}} = g$$, and at height $$h$$ above the surface, $$g_h = g\left(\frac{R_E}{R_E + h}\right)^2$$.

The ratio of time periods gives us $$\frac{T_2}{T_1} = \sqrt{\frac{g}{g_h}} = \frac{R_E + h}{R_E}$$.

Substituting $$T_1 = 4$$ s and $$T_2 = 6$$ s, we get $$\frac{6}{4} = \frac{R_E + h}{R_E}$$, which gives $$\frac{3}{2} = \frac{R_E + h}{R_E}$$.

So $$R_E + h = \frac{3}{2}R_E$$, which means $$h = \frac{1}{2}R_E = \frac{1}{2} \times 6400 = 3200$$ km.

Hence, the correct answer is Option C.

We need to find the percentage change in acceleration due to gravity when the radius of Earth shrinks by 2% while mass remains constant. Using the relation

$$g = \frac{GM}{R^2}$$

and taking its differential yields

$$\frac{\Delta g}{g} = -2\frac{\Delta R}{R}.$$

Since the radius decreases by 2%, i.e., $$\frac{\Delta R}{R} = -2\%,$$ it follows that

$$\frac{\Delta g}{g} = -2 \times (-2\%) = +4\%.$$

Equivalently, because $$g \propto \frac{1}{R^2},$$ a reduction in R increases g, so a 2% decrease in the radius brings about approximately a 4% increase in g. Hence, the correct answer is Option D: increase by 4%.

We need to find the acceleration due to gravity at point $$P$$ which is at a height equal to the diameter of the Earth above the surface. The height of point $$P$$ above the surface is the diameter of the Earth, $$2R$$, where $$R$$ is the radius of the Earth, so the distance from the center of the Earth is $$d = R + h = R + 2R = 3R$$.

The acceleration due to gravity at a distance $$d$$ from the center (when $$d > R$$) is given by $$g' = \frac{g R^2}{d^2}$$. Substituting $$d = 3R$$ yields $$g' = \frac{g R^2}{(3R)^2} = \frac{g R^2}{9R^2} = \frac{g}{9}$$.

The correct answer is Option D: $$\frac{g}{9}$$.

We need to find the percentage decrease in weight when a rocket is taken to a height of $$h = 32 \text{ km}$$ above the Earth's surface. The radius of Earth is $$R = 6400 \text{ km}$$.

We start by expressing the acceleration due to gravity at height $$h$$ above the surface as:

$$g' = g\left(\frac{R}{R + h}\right)^2$$

Since $$h \ll R$$ ($$32 \text{ km} \ll 6400 \text{ km}$$), we can use the binomial approximation:

$$g' \approx g\left(1 - \frac{2h}{R}\right)$$

Next, the fractional decrease in gravitational acceleration is given by:

$$\frac{\Delta g}{g} = \frac{g - g'}{g} = \frac{2h}{R}$$

Substituting the values, we get $$\frac{\Delta g}{g} = \frac{2 \times 32}{6400} = \frac{64}{6400} = \frac{1}{100} = 0.01$$

Finally, converting this to a percentage gives:

$$\text{Percentage decrease} = \frac{\Delta g}{g} \times 100 = 0.01 \times 100 = 1\%$$

Since weight is directly proportional to $$g$$, the percentage decrease in weight is also $$1\%$$.

The correct answer is Option A: $$1\%$$.

We need to find the height $$h$$ from the surface of the Earth where the weight becomes $$\frac{1}{3}$$ of its weight on the surface.

Write the formula for gravitational acceleration at height $$h$$: $$ g' = \frac{g}{\left(1 + \frac{h}{R}\right)^2} $$

Set up the equation: for weight to become $$\frac{1}{3}$$:

$$ \frac{g}{3} = \frac{g}{\left(1 + \frac{h}{R}\right)^2} $$

$$ \left(1 + \frac{h}{R}\right)^2 = 3 $$

$$ 1 + \frac{h}{R} = \sqrt{3} $$

Solve for $$h$$: $$ \frac{h}{R} = \sqrt{3} - 1 $$

$$ h = R(\sqrt{3} - 1) = 6400 \times (1.732 - 1) $$

$$ h = 6400 \times 0.732 = 4684.8 \approx 4685 \text{ km} $$

Therefore, the correct answer is Option B.

We need to identify the correct graph showing the variation of acceleration due to gravity $$g$$ with distance $$r$$ from the center of the Earth.

Since inside a uniform solid sphere the acceleration due to gravity varies linearly with distance from the center, $$g = \frac{4}{3}\pi G \rho \cdot r$$, so $$g \propto r$$ (linear increase from 0 at the center to maximum at the surface).

At the surface ($$r = R$$), $$g = \frac{GM}{R^2}$$ represents the maximum value of $$g$$.

For $$r > R$$, outside the Earth, gravity follows the inverse square law, $$g = \frac{GM}{r^2}$$, therefore $$g \propto \frac{1}{r^2}$$, indicating a decrease.

Combining these results, the graph increases linearly from $$r = 0$$ to $$r = R$$, reaches a peak at $$r = R$$, and then decreases as $$1/r^2$$ for $$r > R$$, resulting in a sharp peak at the surface.

The correct answer is Option A.

Three identical particles A, B, and C each of mass $$100 \text{ kg}$$ are placed in a straight line with $$AB = BC = 13 \text{ m}$$. Particle P of the same mass is placed at $$13 \text{ m}$$ from B on the perpendicular bisector of AC.

We place B at the origin, A at $$(-13, 0)$$, C at $$(13, 0)$$, and P at $$(0, 13)$$.

The distances from P to each particle are found to be $$BP = 13 \text{ m}$$, $$AP = \sqrt{13^2 + 13^2} = 13\sqrt{2} \text{ m}$$, and $$CP = \sqrt{13^2 + 13^2} = 13\sqrt{2} \text{ m}$$.

The gravitational force exerted on P by B is given by $$F_B = \frac{G \times 100 \times 100}{13^2} = \frac{10000G}{169}$$, which acts along BP (vertically upward in this setup).

The gravitational force exerted on P by A is $$F_A = \frac{G \times 100 \times 100}{(13\sqrt{2})^2} = \frac{10000G}{338}$$.

Since AP makes an angle of $$45°$$ with BP, the component of this force along BP is $$F_A \cos 45° = \frac{10000G}{338} \times \frac{1}{\sqrt{2}} = \frac{10000G}{338\sqrt{2}}$$.

By symmetry, the force from C has the same component along BP, and the horizontal components of forces from A and C cancel out, so the net force along BP is: $$F = F_B + 2F_A\cos 45°$$ $$F = \frac{10000G}{169} + 2 \times \frac{10000G}{338\sqrt{2}}$$ $$F = \frac{10000G}{169} + \frac{10000G}{169\sqrt{2}}$$ $$F = \frac{10000G}{169}\left(1 + \frac{1}{\sqrt{2}}\right)$$ $$F = \frac{10000G}{169} \times 1.707$$ $$F \approx 59.2 \times 1.707 \times G$$ $$F \approx 101G \approx 100G$$

Hence, the correct answer is Option B.

Two objects of equal mass $$m$$ are placed at distance $$d$$ apart. The initial gravitational force is $$F = \frac{Gm^2}{d^2}$$.

Find the new masses after transferring one-third of one object's mass.

One-third mass of one object ($$m/3$$) is transferred to the other.

New mass of first object: $$m - \frac{m}{3} = \frac{2m}{3}$$

New mass of second object: $$m + \frac{m}{3} = \frac{4m}{3}$$

Calculate the new gravitational force.

$$F' = \frac{G \times \frac{2m}{3} \times \frac{4m}{3}}{d^2} = \frac{G \times \frac{8m^2}{9}}{d^2} = \frac{8}{9} \times \frac{Gm^2}{d^2} = \frac{8}{9}F$$

The correct answer is Option C.

Statement I: The law of gravitation holds good for any pair of bodies in the universe.

Newton's law of universal gravitation states that every particle in the universe attracts every other particle with a force proportional to the product of their masses and inversely proportional to the square of the distance between them. This is a universal law and holds for any pair of bodies. So, Statement I is true.

Statement II: The weight of any person becomes zero when the person is at the centre of the earth.

At the centre of the earth, the gravitational field is zero because the gravitational pull from all directions cancels out. The acceleration due to gravity at the centre is $$g = 0$$.

Since weight $$W = mg$$, and $$g = 0$$ at the centre:

$$W = m \times 0 = 0$$

So, Statement II is also true.

Both statements are true.

The correct answer is Option A.

We need to find the value of $$x$$ given that the ratio of the lengths $$OA : AB$$ is expressed as $$x : 5$$.

1. Identify the Given Parameters

From the problem :

• Radius of the Earth ($$R$$) = $$6400\text{ km}$$
• Height of point $$C$$ above the surface ($$h$$) = $$3200\text{ km}$$
• Point $$O$$ represents the center of the Earth.
• Point $$B$$ lies on the surface of the Earth, meaning $$OB = R = 6400\text{ km}$$.
• Point $$A$$ is an internal point at a depth $$d$$ below the surface, or at a distance $$r = OA$$ from the center.

2. Express Acceleration Due to Gravity at Points $$A$$ and $$C$$

• At point $$C$$ (at a height $$h$$ above the surface):
  The formula for acceleration due to gravity at an altitude is:

  $$g_C = \frac{g_0 R^2}{(R + h)^2}$$

  Substitute the given values for $$R$$ and $$h$$:

  $$g_C = \frac{g_0 (6400)^2}{(6400 + 3200)^2} = \frac{g_0 (6400)^2}{(9600)^2} = g_0 \left(\frac{2}{3}\right)^2 = \frac{4}{9}g_0$$

  (where $$g_0$$ is the acceleration due to gravity on the Earth's surface)


• At point $$A$$ (at a distance $$r = OA$$ from the center):
  The formula for acceleration due to gravity inside a solid uniform sphere is:

  $$g_A = \frac{g_0 \cdot r}{R}$$

3. Equate the Gravitational Fields to Find $$OA$$

The problem states that the acceleration due to gravity is identical at points $$A$$ and $$C$$ ($$g_A = g_C$$):

$$\frac{g_0 \cdot r}{R} = \frac{4}{9}g_0$$

Cancel out $$g_0$$ from both sides to solve for $$r$$ (which represents the length $$OA$$):

$$r = OA = \frac{4}{9}R$$

4. Determine the Ratio $$OA : AB$$

Since point $$B$$ lies on the surface, the total distance $$OB = OA + AB = R$$. Let's solve for the segment length $$AB$$:

$$AB = R - OA = R - \frac{4}{9}R = \frac{5}{9}R$$

Now, calculate the ratio of the lengths $$OA : AB$$:

$$\frac{OA}{AB} = \frac{\frac{4}{9}R}{\frac{5}{9}R} = \frac{4}{5}$$

We are given that this ratio is equal to $$x : 5$$:

$$\frac{x}{5} = \frac{4}{5} \implies x = 4$$

Conclusion

The value of $$x$$ is 4.

We need to find the compressed radius $$R'$$ of the Earth such that the escape velocity becomes 10 times its current value, while the mass remains the same.

The escape velocity from the surface of a planet is given by $$v_e = \sqrt{\frac{2GM}{R}}$$, where $$M$$ is the mass and $$R$$ is the radius.

Let the new escape velocity be $$v_e' = 10 v_e$$. Since $$v_e' = \sqrt{\frac{2GM}{R'}}$$ and $$v_e = \sqrt{\frac{2GM}{R}}$$, we get $$\frac{v_e'}{v_e} = \sqrt{\frac{R}{R'}}$$.

Substituting $$v_e' = 10 v_e$$, we have $$10 = \sqrt{\frac{R}{R'}}$$, which gives $$100 = \frac{R}{R'}$$, so $$R' = \frac{R}{100}$$.

With $$R = 6400$$ km, the compressed radius is $$R' = \frac{6400}{100} = 64$$ km.

Therefore, the required compressed radius is $$\boxed{64}$$ km.

We need to find the value of $$x$$ to the nearest integer, which represents the numerical coefficient in the expression for the energy required to completely break up the Earth by removing all of its mass to infinity.

Breaking up a solid celestial body completely means separating all of its constituent particles by an infinite distance. The energy required to accomplish this is equal to the magnitude of the gravitational self-energy (or binding energy) of the body.

1. Gravitational Self-Energy of a Uniform Sphere

Assuming the Earth is a perfectly uniform solid sphere of mass $$M$$ and radius $$R$$, its gravitational binding energy ($$U$$) can be derived by integrating the work required to assemble the sphere layer by layer from infinity:

$$U = -\frac{3}{5}\frac{GM^2}{R}$$

The negative sign indicates that the system is bound together by attractive gravitational forces.

2. Energy Required to Disassemble the Earth

To break up the Earth completely and remove all its mass to infinity, we must supply an external energy ($$E$$) equal to the absolute value of this binding energy to bring the total energy of the system up to zero:

$$E = |U| = \frac{3}{5}\frac{GM^2}{R}$$

3. Comparing with the Given Expression

The problem states that the amount of energy that needs to be supplied is given by the expression:

$$E = \frac{x}{5}\frac{GM^2}{R}$$

By comparing our derived formula with the given expression:

$$\frac{3}{5}\frac{GM^2}{R} = \frac{x}{5}\frac{GM^2}{R}$$

This directly simplifies to yield $$x = 3$$.

Therefore, the value of $$x$$ rounded to the nearest integer is 3.

Solution & Explanation

1. Calculate the Initial Gravitational Force ($$F_1$$)

Let the total uniform mass of the initial complete, solid sphere of radius $$R$$ be $$M$$. The particle of mass $$m$$ is placed at a point $$A$$, which is at a distance of $$3R$$ from the center $$O$$ of the sphere.

Using Newton's Law of Universal Gravitation, the initial force $$F_1$$ exerted by the complete sphere on the particle is:

$$F_1 = \frac{G \cdot M \cdot m}{(3R)^2} = \frac{G \cdot M \cdot m}{9R^2} \quad \text{--- (Eq. 1)}$$

2. Find the Mass and Position of the Cavity ($$M_c$$)

The volume of a sphere is proportional to the cube of its radius ($$V \propto R^3$$). Since the sphere has a uniform mass density, we can calculate the mass of the removed cavity ($$M_c$$) of radius $$R_c = \frac{R}{2}$$ as follows:

$$M_c = M \times \left(\frac{R/2}{R}\right)^3 = M \times \left(\frac{1}{2}\right)^3 = \frac{M}{8}$$

From the geometry shown in the figure:

• The center of the cavity $$B$$ is shifted to the right from the center $$O$$ by a distance equal to its radius: $$OB = \frac{R}{2}$$.
• The particle at $$A$$ is at a distance of $$OA = 3R$$ from the main center $$O$$.
• Therefore, the distance from the center of the cavity $$B$$ to the particle $$A$$ is:

  $$BA = OA - OB = 3R - \frac{R}{2} = \frac{5R}{2}$$

3. Calculate the New Gravitational Force ($$F_2$$)

Using the principle of superposition, the net gravitational force exerted by the remaining body with the cavity ($$F_2$$) is equal to the force of the original complete sphere minus the force that would be exerted by the removed cavity mass:

$$F_2 = F_1 - F_c$$

Let's calculate the gravitational pull ($$F_c$$) of the cavity section independently:

$$F_c = \frac{G \cdot M_c \cdot m}{(BA)^2} = \frac{G \cdot \left(\frac{M}{8}\right) \cdot m}{\left(\frac{5R}{2}\right)^2} = \frac{G \cdot M \cdot m}{8 \times \frac{25R^2}{4}} = \frac{G \cdot M \cdot m}{50R^2}$$

Now, subtract this value from $$F_1$$ to determine $$F_2$$:

$$F_2 = \frac{G \cdot M \cdot m}{9R^2} - \frac{G \cdot M \cdot m}{50R^2}$$

$$F_2 = \frac{G \cdot M \cdot m}{R^2} \left( \frac{1}{9} - \frac{1}{50} \right) = \frac{G \cdot M \cdot m}{R^2} \left( \frac{50 - 9}{450} \right) = \frac{41}{450} \cdot \frac{G \cdot M \cdot m}{R^2} \quad \text{--- (Eq. 2)}$$

4. Compute the Ratio $$F_1 : F_2$$

Divide Equation 1 by Equation 2 to obtain the direct numerical proportion:

$$\frac{F_1}{F_2} = \frac{\frac{1}{9}}{\frac{41}{450}} = \frac{1}{9} \times \frac{450}{41} = \frac{50}{41}$$

Thus, the ratio of the initial force to the final remaining force is exactly $$50 : 41$$.

Correct Option: B ($$50 : 41$$)

We recall the shell theorem, which states two key results for a thin, uniform, spherical shell of total mass $$M$$ and radius $$R$$:

1. For any internal point whose distance from the centre is $$r<R$$, the gravitational field (intensity) is zero:

$$\vec g(r)=0.$$

2. The gravitational potential at an internal point equals the potential at the surface and is therefore a constant given by

$$V(r)=V(R)=-\dfrac{GM}{R},$$

where $$G$$ is the universal gravitational constant. The minus sign shows the usual convention that potential is taken to be zero at infinity.

Now we examine each statement in the question:

    (a) “The gravitational field is zero.”  We have just established $$\vec g(r)=0,$$ so statement (a) is true.

    (b) “The gravitational potential is zero.”  From $$V(r)=-\dfrac{GM}{R},$$ the value is generally a non-zero negative constant (unless the mass itself is zero). Hence statement (b) is false.

    (c) “The gravitational field is the same everywhere.”  Because the field is identically zero at every internal point, every location experiences the same value (namely zero). So statement (c) is true.

    (d) “The gravitational potential is the same everywhere.”  We have $$V(r)=$$ constant for all internal points, so statement (d) is true.

    (e) “All the above.”  Since statement (b) is false, “all” cannot be correct, so (e) is false.

Combining the valid statements, only (a), (c) and (d) are correct. These correspond to Option A.

Hence, the correct answer is Option A.

We are given a comet orbiting the Sun with maximum distance (aphelion) $$r_{\text{max}} = 1.6 \times 10^{12} \text{ m}$$, minimum distance (perihelion) $$r_{\text{min}} = 8.0 \times 10^{10} \text{ m}$$, and speed at the nearest point $$v_{\text{near}} = 6 \times 10^4 \text{ m/s}$$.

By conservation of angular momentum about the Sun, at the nearest and farthest points the velocity is perpendicular to the radius vector, so $$m \, v_{\text{near}} \, r_{\text{min}} = m \, v_{\text{far}} \, r_{\text{max}}$$.

Solving for $$v_{\text{far}}$$: $$v_{\text{far}} = \frac{v_{\text{near}} \, r_{\text{min}}}{r_{\text{max}}} = \frac{6 \times 10^4 \times 8.0 \times 10^{10}}{1.6 \times 10^{12}}$$.

Computing: $$v_{\text{far}} = \frac{48 \times 10^{14}}{1.6 \times 10^{12}} = 30 \times 10^{2} = 3.0 \times 10^3 \text{ m/s}$$.

The escape velocity from the surface of a planet of mass $$M$$ and radius $$R$$ is given by $$v_e = \sqrt{\frac{2GM}{R}}$$.

For the escape velocities of planets A and B to be equal, we need $$\frac{M_1}{R_1} = \frac{M_2}{R_2}$$. This condition allows planets of unequal mass to have the same escape velocity, as long as the ratio of mass to radius is the same. For example, if planet A has twice the mass and twice the radius of planet B, they will have the same escape velocity. Therefore, Assertion A is correct.

The Reason states that $$M_1R_1 = M_2R_2$$. However, from the escape velocity formula, the actual requirement is $$\frac{M_1}{R_1} = \frac{M_2}{R_2}$$, which gives $$M_1R_2 = M_2R_1$$, not $$M_1R_1 = M_2R_2$$. The condition given in Reason R is incorrect.

Therefore, A is correct but R is not correct, which corresponds to Option (4).

We have to find the ratio between the acceleration due to gravity at a depth $$r$$ below the Earth’s surface and at a height $$r$$ above the surface, where $$r \lt R_E$$ and $$R_E$$ is the Earth’s radius.

First recall the standard results for the variation of acceleration due to gravity with depth and with height:

1. Variation with depth. For a depth $$r$$ (or $$d$$) inside the Earth, the formula is stated as $$g_d = g\left(1 - \frac{r}{R_E}\right)$$ because only the mass enclosed within radius $$R_E - r$$ contributes to the gravitational pull, leading to a linear decrease.

2. Variation with height. For a height $$r$$ (or $$h$$) above the surface, the formula is stated as $$g_h = \frac{g}{\left(1 + \frac{r}{R_E}\right)^2}$$ because the field outside the Earth follows the inverse-square law while the distance from the centre becomes $$R_E + r$$.

Now we are asked for the ratio $$\dfrac{g_d}{g_h}$$. Substituting the two formulae, we write

$$ \frac{g_d}{g_h} = \frac{g\left(1 - \frac{r}{R_E}\right)} {\displaystyle \frac{g}{\left(1 + \frac{r}{R_E}\right)^2}} = \left(1 - \frac{r}{R_E}\right)\left(1 + \frac{r}{R_E}\right)^2 . $$

The factor $$g$$ cancels out, as expected. To make the algebra clearer, set

$$x = \frac{r}{R_E}, \qquad \text{with } x \lt 1.$$

With this substitution the ratio becomes

$$ \frac{g_d}{g_h} = (1 - x)(1 + x)^2 . $$

Now expand every factor step by step. First expand the square:

$$ (1 + x)^2 = 1 + 2x + x^2 . $$

Next multiply this result by the remaining factor $$(1 - x)$$:

$$ (1 - x)(1 + 2x + x^2) = 1 + 2x + x^2 \;-\; x - 2x^2 - x^3 . $$

Combine like terms:

$$ 1 + 2x + x^2 - x - 2x^2 - x^3 = 1 + (2x - x) + (x^2 - 2x^2) - x^3 = 1 + x - x^2 - x^3 . $$

Finally replace $$x$$ by $$\dfrac{r}{R_E}$$ to return to the original variables:

$$ \frac{g_d}{g_h} = 1 + \frac{r}{R_E} - \frac{r^2}{R_E^2} - \frac{r^3}{R_E^3}. $$

This result matches exactly the expression given in Option A.

Hence, the correct answer is Option A.

We have two stars of masses $$m$$ and $$2m$$ separated by a distance $$d$$, rotating about their common centre of mass.

The centre of mass divides the line joining them in the inverse ratio of their masses. So the distance of mass $$m$$ from the centre of mass is $$r_1 = \frac{2m \cdot d}{m + 2m} = \frac{2d}{3}$$, and the distance of mass $$2m$$ from the centre of mass is $$r_2 = \frac{m \cdot d}{3m} = \frac{d}{3}$$.

The gravitational force between them provides the centripetal force. For the star of mass $$m$$ orbiting at radius $$r_1 = \frac{2d}{3}$$, we write $$m\omega^2 r_1 = \frac{G \cdot m \cdot 2m}{d^2}$$.

Substituting $$r_1 = \frac{2d}{3}$$, we get $$m\omega^2 \cdot \frac{2d}{3} = \frac{2Gm^2}{d^2}$$.

Simplifying, $$\omega^2 = \frac{2Gm}{d^2} \times \frac{3}{2d} = \frac{3Gm}{d^3}$$.

The period of revolution is $$T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{d^3}{3Gm}}$$.

Hence, the correct answer is Option A.

First, recall that the weight of an object on any planet is given by the familiar relation $$W = m\,g,$$ where $$m$$ is the mass of the object and $$g$$ is the acceleration due to gravity on that planet. For two different planets we can compare the weights by comparing their respective values of $$g.$$

The acceleration due to gravity at the surface of a spherical planet is obtained from Newton’s law of gravitation:

$$g = \dfrac{G\,M}{R^{2}},$$

where $$G$$ is the universal gravitational constant, $$M$$ is the mass of the planet, and $$R$$ is its radius. Thus, if we know how the mass and radius of another planet relate to the Earth’s, we can get the ratio of the two $$g$$’s.

We are told that the new planet has a mass which is double the mass of the Earth. Writing Earth’s mass as $$M_E,$$ we have

$$M_P = 2\,M_E.$$

Next, we are told that the density of the new planet is equal to the average density of the Earth. Density is defined as mass divided by volume. For a sphere, volume is $$\dfrac{4}{3}\pi R^{3},$$ so

$$\rho = \dfrac{M}{\frac{4}{3}\pi R^{3}} \quad\Longrightarrow\quad M = \rho\,\frac{4}{3}\pi R^{3}.$$

If the densities of the two planets are equal, we can set up the equality

$$\rho_P = \rho_E \quad\Longrightarrow\quad \dfrac{M_P}{\frac{4}{3}\pi R_P^{3}} = \dfrac{M_E}{\frac{4}{3}\pi R_E^{3}}.$$

Cancelling the common factors $$\frac{4}{3}\pi$$ on both sides gives

$$\dfrac{M_P}{R_P^{3}} = \dfrac{M_E}{R_E^{3}}.$$

Substituting $$M_P = 2\,M_E$$ into the above relation, we get

$$\dfrac{2\,M_E}{R_P^{3}} = \dfrac{M_E}{R_E^{3}}.$$

Dividing both sides by $$M_E$$ and then cross-multiplying, we have

$$2\,R_E^{3} = R_P^{3}.$$

Taking the cube root of both sides, we obtain the radius of the planet in terms of Earth’s radius:

$$R_P = 2^{\frac{1}{3}}\,R_E.$$

Now we can find the ratio of the gravitational accelerations. Using the formula $$g = \dfrac{G\,M}{R^{2}},$$ we write

$$\dfrac{g_P}{g_E} = \dfrac{\dfrac{G\,M_P}{R_P^{2}}}{\dfrac{G\,M_E}{R_E^{2}}}.$$

The constant $$G$$ cancels out, so

$$\dfrac{g_P}{g_E} = \dfrac{M_P}{M_E}\,\dfrac{R_E^{2}}{R_P^{2}}.$$

Substituting $$M_P = 2\,M_E$$ and $$R_P = 2^{\frac{1}{3}}\,R_E,$$ we get

$$\dfrac{g_P}{g_E} = 2 \times \dfrac{R_E^{2}}{\left(2^{\frac{1}{3}}\,R_E\right)^{2}}.$$

Simplifying the expression inside the denominator first,

$$(2^{\tfrac{1}{3}}\,R_E)^{2} = 2^{\tfrac{2}{3}}\,R_E^{2}.$$

Therefore,

$$\dfrac{g_P}{g_E} = 2 \times \dfrac{R_E^{2}}{2^{\tfrac{2}{3}}\,R_E^{2}}.$$

The factor $$R_E^{2}$$ cancels, giving

$$\dfrac{g_P}{g_E} = \dfrac{2}{2^{\tfrac{2}{3}}} = 2^{1 - \tfrac{2}{3}} = 2^{\tfrac{1}{3}}.$$

Finally, the weight of the object on the new planet will be

$$W_P = m\,g_P = m\,g_E \left(\dfrac{g_P}{g_E}\right) = W \times 2^{\tfrac{1}{3}}.$$

So the weight becomes $$2^{\tfrac{1}{3}}\,W.$$

Hence, the correct answer is Option C.

We are told that one of Mars’ moons has an orbital period of 7 hours 30 minutes and an orbital radius of $$9.0 \times 10^3$$ km. First we convert all the given data into SI units.

The period is 7 hours 30 minutes. Since 30 minutes is half an hour, this is a total of 7.5 hours. Converting hours to seconds we use the relation $$1 \text{ hour} = 3600 \text{ s}.$$ So $$T = 7.5 \times 3600 \text{ s} = 27000 \text{ s} = 2.7 \times 10^4 \text{ s}.$$

The radius is $$9.0 \times 10^3$$ km. Knowing that $$1 \text{ km} = 1000 \text{ m},$$ we get $$r = 9.0 \times 10^3 \times 10^3 \text{ m} = 9.0 \times 10^6 \text{ m}.$$

For a satellite orbiting a planet, Kepler’s third law (derived from Newton’s law of gravitation) states $$T^2 = \frac{4\pi^2}{G M}\,r^3,$$ where $$M$$ is the mass of the planet. We are asked to find $$M$$, so we solve this equation for $$M$$. Multiplying both sides by $$G M$$ and then dividing by $$T^2$$ gives

$$M = \frac{4\pi^2 r^3}{G T^2}.$$

The problem supplies the convenient constant $$\frac{4\pi^2}{G} = 6 \times 10^{11}\;\text{N}^{-1}\text{ m}^{-2}\text{ kg}^2.$$ Using this, we can rewrite the expression for the mass as

$$M = \left(\frac{4\pi^2}{G}\right)\frac{r^3}{T^2}.$$

Now we compute each part step by step. First, the cube of the radius:

$$r^3 = \left(9.0 \times 10^6\right)^3 = 9^3 \times 10^{18} = 729 \times 10^{18} = 7.29 \times 10^{20} \text{ m}^3.$$

Next, the square of the period:

$$T^2 = \left(2.7 \times 10^4\right)^2 = 2.7^2 \times 10^{8} = 7.29 \times 10^{8} \text{ s}^2.$$

We now form the ratio $$\dfrac{r^3}{T^2}$$:

$$\frac{r^3}{T^2} = \frac{7.29 \times 10^{20}}{7.29 \times 10^{8}} = 10^{12} \text{ m}^3\text{ s}^{-2}.$$

Finally, substituting this result into the mass formula, we get

$$M = \left(6 \times 10^{11}\right)\left(10^{12}\right) \text{ kg} = 6 \times 10^{23} \text{ kg}.$$

Hence, the correct answer is Option D.

We are asked to find the gravitational potential at a point that lies inside a thin, uniform spherical shell of mass $$100\ \text{kg}$$ and radius $$50\ \text{m}$$, while a point mass of $$50\ \text{kg}$$ sits exactly at the common centre. The point of interest is situated at a distance $$25\ \text{m}$$ from this centre.

We recall two standard results from Newtonian gravitation:

1. Potential due to a point mass. For a point mass $$m$$, the gravitational potential at a distance $$r$$ is given by the formula $$\displaystyle V_{\text{point}} = -\,\frac{G\,m}{r}$$ where $$G$$ is the universal gravitational constant.

2. Potential inside a uniform spherical shell. At any interior point (including the very centre) of a thin spherical shell of total mass $$M$$ and radius $$R$$, the potential has the same constant value $$\displaystyle V_{\text{shell (inside)}} = -\,\frac{G\,M}{R}.$$ It is independent of how deep inside we are, as long as we remain inside the hollow shell.

Because potential is a scalar quantity, the net potential is simply the algebraic sum of the individual potentials contributed by each mass.

We first compute the shell’s contribution. Here $$M = 100\ \text{kg}, \qquad R = 50\ \text{m}.$$ Using the stated formula, we have $$V_{\text{shell}} = -\,\frac{G\,M}{R} = -\,\frac{G \times 100}{50} = -\,\frac{100}{50}\,G = -\,2\,G.$$

Next we compute the point mass’s contribution. For the point mass at the centre, $$m = 50\ \text{kg}, \qquad r = 25\ \text{m}.$$ Using the point-mass formula, $$V_{\text{point}} = -\,\frac{G\,m}{r} = -\,\frac{G \times 50}{25} = -\,\frac{50}{25}\,G = -\,2\,G.$$

Now we add the two potentials:

$$\begin{aligned} V_{\text{total}} & = V_{\text{shell}} + V_{\text{point}} \\ & = (-\,2\,G) + (-\,2\,G) \\ & = -\,4\,G. \end{aligned}$$

Thus the gravitational potential at the given point is

$$V = -\,4\,G.$$

Hence, the correct answer is Option C.

Let the acceleration due to gravity on the surface of the Earth be denoted by $$g$$. We have two different positions to consider for the same body:

1. At a height $$h$$ above the surface. 2. At a depth $$h$$ below the surface.

According to Newton’s law of gravitation, the magnitude of $$g$$ varies with distance from the centre of the Earth. First we state the two standard formulae that connect $$g$$ with height and depth (rotation neglected):

• For a height $$h$$ above the surface, the distance from the centre becomes $$R+h$$.  The formula is $$g_h = g\left(\dfrac{R}{R+h}\right)^2$$.

• For a depth $$h$$ below the surface, only the mass enclosed within the radius $$R-h$$ contributes.  For a uniform Earth, $$g_d = g\left(1-\dfrac{h}{R}\right)$$.

The problem states that the weight (and hence the gravitational acceleration) is the same in both cases, so we equate the two expressions:

$$g\left(\dfrac{R}{R+h}\right)^2 \;=\; g\left(1-\dfrac{h}{R}\right).$$

Because the factor $$g$$ appears on both sides, it cancels out, leaving

$$\left(\dfrac{R}{R+h}\right)^2 = 1-\dfrac{h}{R}.$$

To simplify the algebra, we introduce the dimensionless variable

$$x = \dfrac{h}{R}.$$

Substituting $$h = xR$$ into the equation gives

$$\left(\dfrac{R}{R + xR}\right)^2 = 1 - x.$$

Inside the brackets, factor out $$R$$:

$$\left(\dfrac{R}{R(1 + x)}\right)^2 = 1 - x \;\;\Longrightarrow\;\; \left(\dfrac{1}{1 + x}\right)^2 = 1 - x.$$

Remove the fraction by multiplying both sides by $$(1 + x)^2$$:

$$(1 + x)^2\,(1 - x) = 1.$$

Now expand the left-hand side step by step. First, square the binomial:

$$(1 + x)^2 = 1 + 2x + x^2.$$

Next, multiply this result by $$(1 - x)$$:

$$(1 + 2x + x^2)(1 - x) = 1(1 - x) + 2x(1 - x) + x^2(1 - x).$$

Carrying out each product:

$$1 - x + 2x - 2x^2 + x^2 - x^3.$$

Combine like terms:

$$1 + x - x^2 - x^3.$$

Therefore the equation becomes

$$1 + x - x^2 - x^3 = 1.$$

Subtract $$1$$ from both sides to bring all terms to one side:

$$x - x^2 - x^3 = 0.$$

Factor out the common factor $$x$$:

$$x\bigl(1 - x - x^2\bigr) = 0.$$

One solution is $$x = 0$$, corresponding to $$h = 0$$, which is trivial. For a non-zero height we require

$$1 - x - x^2 = 0.$$

This is a quadratic in $$x$$. Rewriting it in standard form:

$$x^2 + x - 1 = 0.$$

We now apply the quadratic formula $$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ with $$a = 1,\; b = 1,\; c = -1$$:

$$x = \dfrac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)} = \dfrac{-1 \pm \sqrt{1 + 4}}{2} = \dfrac{-1 \pm \sqrt{5}}{2}.$$

The negative root gives a negative height, which is not meaningful here, so we choose the positive root:

$$x = \dfrac{-1 + \sqrt{5}}{2}.$$

Now revert to the original variable $$h$$ using $$x = \dfrac{h}{R}$$:

$$\dfrac{h}{R} = \dfrac{\sqrt{5} - 1}{2} \;\;\Longrightarrow\;\; h = \dfrac{\sqrt{5}\,R - R}{2}.$$

This expression exactly matches Option C.

Hence, the correct answer is Option C.

We have a satellite of mass $$M$$ that is fired vertically from the surface of the earth (radius $$R$$) with speed $$u$$. Its total mechanical energy just after launch is

$$E_0=\frac12Mu^{2}-\frac{GM_eM}{R}.$$

When it rises to a height $$R$$ above the surface its distance from the earth’s centre is $$r=R+R=2R$$. At that instant let its speed be $$v$$. Applying conservation of mechanical energy between the point of launch and the point $$r=2R$$, we write

$$\frac12Mu^{2}-\frac{GM_eM}{R}= \frac12Mv^{2}-\frac{GM_eM}{2R}.$$

Simplifying,

$$\frac12Mu^{2}-\frac{GM_eM}{R}+ \frac{GM_eM}{2R}= \frac12Mv^{2}

\;\Longrightarrow\;

\frac12Mv^{2}= \frac12Mu^{2}-\frac{GM_eM}{2R}.$$

Dividing by $$\tfrac12M$$ we get

$$v^{2}=u^{2}-\frac{GM_e}{R}. \quad -(1)$$

Immediately afterwards the satellite ejects a small rocket of mass $$\dfrac{M}{10}$$, so the remaining mass of the satellite becomes

$$M_s=\frac{9M}{10}.$$

The problem states that the satellite (of mass $$\dfrac{9M}{10}$$) now moves in a circular orbit of radius $$2R$$. The speed required for a circular orbit of radius $$r$$ is obtained from the centripetal-gravitational balance

$$\frac{v_c^{2}}{r}=\frac{GM_e}{r^{2}}

\;\Longrightarrow\;

v_c=\sqrt{\frac{GM_e}{r}}

=\sqrt{\frac{GM_e}{2R}}. \quad -(2)$$

During the extremely short interval of ejection, the only external force is gravity, which is radial; its impulse is therefore negligible. Hence the linear momentum of the system (satellite + rocket) is conserved separately in the radial and tangential directions.

Radial direction. Before ejection the entire mass $$M$$ is moving radially outward with speed $$v$$, so the initial radial momentum is

$$P_{r,\,\text{initial}}=Mv.$$

After ejection, the satellite is moving purely tangentially (no radial component), so all the radial momentum must be carried by the rocket (mass $$\dfrac{M}{10}$$). Let $$v_{r,\parallel}$$ be the radial component of the rocket’s velocity. Conservation gives

$$Mv=\frac{M}{10}\,v_{r,\parallel}

\;\Longrightarrow\;

v_{r,\parallel}=10v. \quad -(3)$$

Tangential direction. Initially the tangential momentum is zero (vertical launch). After ejection the satellite (mass $$\dfrac{9M}{10}$$) moves tangentially with speed $$v_c$$. Let $$v_{r,\perp}$$ be the tangential component of the rocket’s velocity. Conservation of tangential momentum gives

$$0=\frac{9M}{10}\,v_c+\frac{M}{10}\,v_{r,\perp}

\;\Longrightarrow\;

v_{r,\perp}=-9v_c. \quad -(4)$$

(The negative sign shows that the rocket’s tangential motion is opposite to that of the satellite.)

Magnitude of the rocket’s velocity. Because the radial and tangential components are perpendicular, the speed of the rocket is

$$v_r=\sqrt{v_{r,\parallel}^{2}+v_{r,\perp}^{2}}

=\sqrt{(10v)^{2}+(-9v_c)^{2}}

=\sqrt{100v^{2}+81v_c^{2}}. \quad -(5)$$

Kinetic energy of the rocket. Using $$K=\tfrac12mv^{2}$$,

$$K_r=\frac12\left(\frac{M}{10}\right)v_r^{2}

=\frac{M}{20}\bigl(100v^{2}+81v_c^{2}\bigr). \quad -(6)$$

Now substitute from (1) and (2):

$$v^{2}=u^{2}-\frac{GM_e}{R}, \qquad

v_c^{2}=\frac{GM_e}{2R}.$$

Putting these into (6),

$$

\begin{aligned}

K_r

&=\frac{M}{20}\Bigl[100\Bigl(u^{2}-\frac{GM_e}{R}\Bigr)

+81\Bigl(\frac{GM_e}{2R}\Bigr)\Bigr] \\

&=\frac{M}{20}\Bigl[100u^{2}-100\frac{GM_e}{R}

+\frac{81}{2}\frac{GM_e}{R}\Bigr] \\

&=\frac{M}{20}\Bigl[100u^{2}-100\frac{GM_e}{R}

+40.5\frac{GM_e}{R}\Bigr] \\

&=\frac{M}{20}\Bigl[100u^{2}-59.5\frac{GM_e}{R}\Bigr].

\end{aligned}

$$

Factorising the common factor $$100$$ inside the bracket,

$$

K_r

=\frac{M}{20}\cdot100\Bigl[u^{2}-0.595\frac{GM_e}{R}\Bigr]

=5M\left(u^{2}-\frac{119}{200}\frac{GM_e}{R}\right). \quad -(7)

$$

This matches Option B exactly.

Hence, the correct answer is Option 2.

Minimum Required Theory

• Gravitational Field of a Solid Sphere: The gravitational field intensity ($$E$$) at a distance $$r$$ from the center of a solid sphere of mass $$M$$ and radius $$R$$ follows two regimes:
  • Inside ($$r < R$$): $$E \propto r$$ (linear increase)
• Outside ($$r \ge R$$): $$E = \frac{GM}{r^2}$$ (inverse-square decrease)
• Peak Value: The maximum gravitational field occurs exactly at the surface of the sphere ($$r = R$$), where:
• Radius $$R_1 = 1\text{ m}$$
• Peak gravitational field value at its surface ($$r = 1\text{ m}$$) is $$E_1 = 2$$
• Radius $$R_2 = 2\text{ m}$$
• Peak gravitational field value at its surface ($$r = 2\text{ m}$$) is $$E_2 = 3$$

$$E_{max} = \frac{GM}{R^2}$$

Step-by-Step Solution

Step 1: Extract data from the graph for Sphere (1)

Using the surface field formula:

$$2 = \frac{GM_1}{(1)^2} \implies 2 = GM_1 \quad \text{--- (Equation 1)}$$

Step 2: Extract data from the graph for Sphere (2)

Using the surface field formula:

$$3 = \frac{GM_2}{(2)^2} \implies 3 = \frac{GM_2}{4} \implies 12 = GM_2 \quad \text{--- (Equation 2)}$$

Step 3: Calculate the ratio $$\frac{M_1}{M_2}$$

Divide Equation 1 by Equation 2:

$$\frac{GM_1}{GM_2} = \frac{2}{12}$$

Cancel the universal gravitational constant $$G$$:

$$\frac{M_1}{M_2} = \frac{1}{6}$$

Final Answer

The value of $$\frac{M_1}{M_2}$$ is $$\frac{1}{6}$$ (or $$0.167$$).

We are told that on the $$x$$-axis the gravitational field (which we may denote by $$\vec g(x)$$) has only an $$x$$-component and is given by

$$\vec g(x)=\frac{A x}{(x^2+a^2)^{3/2}}\;\hat i.$$

For a one-dimensional situation along the $$x$$-axis, the relation between gravitational field and gravitational potential $$\phi(x)$$ is the well-known formula

$$\vec g(x)= -\,\frac{d\phi}{dx}\;\hat i.$$

In scalar form (the direction is already along $$\hat i$$) this becomes

$$\frac{d\phi}{dx}= -\,\frac{A x}{(x^2+a^2)^{3/2}}.$$

We now integrate both sides with respect to $$x$$. Taking the potential to be zero at infinity, we write

$$\phi(x)-\phi(\infty)=\int_{\infty}^{x}\frac{d\phi}{dx'}\,dx'=-\int_{\infty}^{x}\frac{A x'}{(x'^2+a^2)^{3/2}}\,dx'.$$

Because $$\phi(\infty)=0$$, this simplifies to

$$\phi(x)= -A\int_{\infty}^{x}\frac{x'}{(x'^2+a^2)^{3/2}}\;dx'.$$

To evaluate the integral, we first consider the indefinite form

$$\int \frac{x}{(x^2+a^2)^{3/2}}\;dx.$$

We observe that

$$\frac{d}{dx}\left(\frac{1}{\sqrt{x^2+a^2}}\right)=\frac{d}{dx}\left((x^2+a^2)^{-1/2}\right)= -\frac{1}{2}(x^2+a^2)^{-3/2}\cdot 2x=-\frac{x}{(x^2+a^2)^{3/2}}.$$

Thus,

$$\int \frac{x}{(x^2+a^2)^{3/2}}\;dx = -\frac{1}{\sqrt{x^2+a^2}}+C.$$

Substituting this result back into the definite integral we have

$$\phi(x)= -A\Bigl[-\frac{1}{\sqrt{x'^2+a^2}}\Bigr]_{\infty}^{x}=A\Bigl[\frac{1}{\sqrt{x'^2+a^2}}\Bigr]_{\infty}^{x}.$$

Now we evaluate the limits. At the upper limit $$x' = x$$ we obtain $$\dfrac{1}{\sqrt{x^2+a^2}}$$, while at the lower limit $$x' \to \infty$$ we have $$\dfrac{1}{\sqrt{\infty}}=0$$. Hence

$$\phi(x)=\frac{A}{\sqrt{x^2+a^2}}.$$

Therefore the magnitude of the gravitational potential on the $$x$$-axis at a distance $$x$$ is

$$\boxed{\displaystyle \frac{A}{\sqrt{x^2+a^2}}}.$$

Hence, the correct answer is Option A.

Let $$g$$ denote the acceleration due to gravity at the surface of the Earth. We need the expressions for gravity when we go up to a height and when we go down to a depth.

First, the standard relation for acceleration due to gravity at a height $$h$$ above the surface is stated as $$g_h = g\left(\frac{R}{R+h}\right)^2.$$ This follows from Newton’s law of gravitation, because the distance of the mass from the centre becomes $$R+h$$ and the force is inversely proportional to the square of that distance.

Second, the standard relation for acceleration due to gravity at a depth $$d$$ below the surface is stated as $$g_d = g\left(1-\frac{d}{R}\right).$$ Inside the Earth the gravitational field falls off linearly with distance from the centre, giving this simple linear reduction factor.

According to the question we are given that the value of gravity at height $$h = \dfrac{R}{2}$$ equals the same value at some depth $$d$$. So we write $$g_h = g_d.$$

Substituting the formulae, we have $$g\left(\frac{R}{R+\dfrac{R}{2}}\right)^2 = g\left(1-\frac{d}{R}\right).$$

Now we simplify the left-hand side step by step. The denominator inside the parentheses is $$R + \frac{R}{2} = \frac{3R}{2}.$$ So inside the square we get $$\frac{R}{\dfrac{3R}{2}} = \frac{R}{1}\times \frac{2}{3R} = \frac{2}{3}.$$

Squaring this fraction, $$\left(\frac{2}{3}\right)^2 = \frac{4}{9}.$$ Hence the left-hand side becomes $$g \times \frac{4}{9}.$$

The equation thus reduces to $$g\left(\frac{4}{9}\right) = g\left(1-\frac{d}{R}\right).$$

We can cancel the common factor $$g$$ on both sides, giving $$\frac{4}{9} = 1-\frac{d}{R}.$$

Now we isolate the ratio $$\dfrac{d}{R}$$. Transposing the fraction yields $$\frac{d}{R} = 1 - \frac{4}{9}.$$

Writing the right-hand side with a common denominator, $$1 = \frac{9}{9},$$ so $$\frac{d}{R} = \frac{9}{9} - \frac{4}{9} = \frac{5}{9}.$$

Thus the required ratio is $$\boxed{\dfrac{d}{R} = \dfrac{5}{9}}.$$

Hence, the correct answer is Option B.

We are told that the mass-density inside the planet depends on the distance from the centre according to

$$\rho(r)=\rho_0\!\left(1-\frac{r^{2}}{R^{2}}\right).$$

To find the point where the gravitational field $$g(r)$$ is maximum we must first obtain an explicit expression for $$g(r)$$ inside the planet and then differentiate it with respect to $$r$$.

The gravitational field at a point situated at a distance $$r<R$$ from the centre depends only on the mass contained within the sphere of radius $$r$$. By Gauss’s law for gravitation (equivalent to Newton’s law), we have

$$g(r)=\frac{G\,M(r)}{r^{2}},$$

where $$M(r)$$ is the mass enclosed within radius $$r$$.

To calculate $$M(r)$$, we integrate the density over the volume from the centre up to $$r$$:

$$M(r)=\int_{0}^{r} 4\pi r'^{2}\,\rho(r')\,dr'.$$

Substituting $$\rho(r')=\rho_0\!\left(1-\dfrac{r'^{2}}{R^{2}}\right)$$, we get

$$M(r)=4\pi\rho_0\!\int_{0}^{r}\!\left(1-\frac{r'^{2}}{R^{2}}\right)r'^{2}dr'.$$

Expanding inside the integral,

$$M(r)=4\pi\rho_0\!\left[\int_{0}^{r}r'^{2}dr'-\frac{1}{R^{2}}\int_{0}^{r}r'^{4}dr'\right].$$

Using the power-integral formulas $$\displaystyle\int r'^{2}dr'=\frac{r'^{3}}{3}$$ and $$\displaystyle\int r'^{4}dr'=\frac{r'^{5}}{5},$$ we obtain

$$M(r)=4\pi\rho_0\!\left[\frac{r^{3}}{3}-\frac{r^{5}}{5R^{2}}\right].$$

Now we substitute this result into $$g(r)=\dfrac{G\,M(r)}{r^{2}}$$:

$$g(r)=\frac{G}{r^{2}}\;4\pi\rho_0\!\left[\frac{r^{3}}{3}-\frac{r^{5}}{5R^{2}}\right].$$

Simplifying the factors of $$r$$,

$$g(r)=4\pi G\rho_0\left[\frac{r}{3}-\frac{r^{3}}{5R^{2}}\right].$$

It is convenient to pull out a common constant $$k=4\pi G\rho_0$$, so

$$g(r)=k\,r\!\left(\frac13-\frac{r^{2}}{5R^{2}}\right).$$

To locate the maximum of $$g(r)$$ for $$0<r<R$$ we differentiate with respect to $$r$$ and set the derivative to zero. Let

$$f(r)=r\!\left(\frac13-\frac{r^{2}}{5R^{2}}\right) \quad\text{so that}\quad g(r)=k\,f(r).$$

Differentiating term by term, we have

$$\frac{df}{dr}=\left(\frac13-\frac{r^{2}}{5R^{2}}\right)+r\!\left(-\frac{2r}{5R^{2}}\right).$$

The two terms give

$$\frac{df}{dr}=\frac13-\frac{r^{2}}{5R^{2}}-\frac{2r^{2}}{5R^{2}}=\frac13-\frac{3r^{2}}{5R^{2}}.$$

Setting $$\dfrac{df}{dr}=0$$ for an extremum,

$$\frac13-\frac{3r^{2}}{5R^{2}}=0.$$ $$\frac{3r^{2}}{5R^{2}}=\frac13.$$ $$r^{2}=R^{2}\left(\frac13\cdot\frac{5}{3}\right)=\frac{5}{9}R^{2}.$$

Taking the positive square root (since $$r$$ is a radial distance),

$$r=\sqrt{\frac{5}{9}}\;R=\frac{\sqrt5}{3}\,R.$$

This value indeed lies within the planet ($$r<R$$) and represents the point where $$g(r)$$ attains its maximum. Comparing with the given choices, this corresponds to Option D.

Hence, the correct answer is Option D.

We know that the acceleration due to gravity at any distance from the centre of Earth is given by the Newtonian formula $$g \;=\; \dfrac{G\,M}{r^{\,2}}$$ where $$G$$ is the gravitational constant, $$M$$ is the mass of Earth and $$r$$ is the distance of the point from the centre of Earth.

At Earth’s surface the distance from the centre is simply the radius $$R$$ of Earth, so the surface value is

$$g \;=\; \dfrac{G\,M}{R^{\,2}} \;=\; 9.8 \text{ m s}^{-2}.$$

If we go to an altitude $$h$$ above the surface, the distance from the centre becomes $$R+h$$ and the acceleration due to gravity there (call it $$g'$$) is

$$g' \;=\; \dfrac{G\,M}{(R+h)^{2}}.$$

To find the relation between $$g'$$ and $$g$$, we divide the two expressions:

$$\dfrac{g'}{g} \;=\; \dfrac{\dfrac{G\,M}{(R+h)^{2}}}{\dfrac{G\,M}{R^{\,2}}} \;=\; \dfrac{R^{\,2}}{(R+h)^{2}}.$$

Simplifying the right-hand side, we get the useful relation

$$g' \;=\; g\;\left(\dfrac{R}{R+h}\right)^{2}.$$

According to the question, the value at altitude must be one-half of the surface value, i.e.

$$g' = 4.9 \text{ m s}^{-2} = \dfrac{1}{2}\,g.$$

Substituting this condition into the previous relation, we have

$$\dfrac{1}{2}\,g \;=\; g\;\left(\dfrac{R}{R+h}\right)^{2}.$$

Now we cancel the common factor $$g$$ on both sides:

$$\dfrac{1}{2} \;=\; \left(\dfrac{R}{R+h}\right)^{2}.$$

Taking the square root of both sides (and keeping only the positive root because distances are positive), we obtain

$$\sqrt{\dfrac{1}{2}} \;=\; \dfrac{R}{R+h},$$

so

$$\dfrac{1}{\sqrt{2}} \;=\; \dfrac{R}{R+h}.$$

Cross-multiplying gives

$$R \;=\; \dfrac{R+h}{\sqrt{2}}.$$

Rewriting this in a clearer form, multiply both sides by $$\sqrt{2}$$:

$$\sqrt{2}\,R \;=\; R + h.$$

Now we isolate $$h$$ by subtracting $$R$$ from both sides:

$$h \;=\; (\sqrt{2} - 1)\,R.$$

We substitute the numerical value of Earth’s radius $$R = 6.4 \times 10^{6} \text{ m}$$ and the approximate value $$\sqrt{2} \approx 1.414$$:

$$h \;=\; (1.414 - 1)\,(6.4 \times 10^{6}) \text{ m}.$$

This difference $$1.414 - 1 = 0.414$$, so

$$h \;=\; 0.414 \times 6.4 \times 10^{6} \text{ m}.$$

Multiplying $$0.414$$ and $$6.4$$ step by step:

$$0.414 \times 6 = 2.484,$$

$$0.414 \times 0.4 = 0.1656,$$

Adding, $$2.484 + 0.1656 = 2.6496.$$

Thus

$$h \approx 2.65 \times 10^{6} \text{ m}.$$

Among the given options the closest numerical value is $$2.6 \times 10^{6} \text{ m}$$, which corresponds to Option B.

Hence, the correct answer is Option B.

We have a point mass of magnitude $$m$$ placed at the origin $$x = 0$$. A thin straight rod lies along the positive $$x$$-axis, beginning at $$x = a$$ and ending at $$x = a + L$$. Its mass per unit length (linear mass density) varies with position according to

$$\lambda(x)=A + Bx^{2}.$$

Our aim is to find the total gravitational force (magnitude) exerted by every small element of the rod on the mass at the origin. Throughout we shall use Newton’s law of gravitation, which in differential form for two small masses separated by a distance $$r$$ is

$$dF=\frac{G\,dm_1\,dm_2}{r^{2}}.$$

Here $$dm_1 = m$$ (the point mass) and $$dm_2 = dm$$ (an infinitesimal element of the rod). The separation between the element located at position $$x$$ and the origin is simply $$r = x$$. Substituting these facts, we get

$$dF = \frac{G\,m\,dm}{x^{2}}.$$

Because each element lies on the positive $$x$$-axis while the point mass is at the origin, every elemental force acts along the negative $$x$$-direction. We shall calculate its magnitude first and later attach the direction if required.

The infinitesimal mass element of the rod is given by the product of its linear density and an infinitesimal length $$dx$$:

$$dm = \lambda(x)\,dx = (A + Bx^{2})\,dx.$$

Placing this expression for $$dm$$ into the formula for $$dF$$, we obtain

$$dF = \frac{G\,m\,(A + Bx^{2})}{x^{2}}\;dx.$$

Now we separate the fraction:

$$dF = Gm\left(\frac{A}{x^{2}} + B\right)\,dx.$$

To find the total force, we integrate $$dF$$ from the near end of the rod at $$x = a$$ to the far end at $$x = a + L$$:

$$F = Gm \int_{a}^{a+L} \left(\frac{A}{x^{2}} + B\right)\,dx.$$

We evaluate the integral term by term.

First term: for $$\displaystyle \int \frac{A}{x^{2}}\,dx$$, we recall the standard result

$$\int x^{-2}\,dx = -x^{-1} + C,$$

so

$$\int \frac{A}{x^{2}}\,dx = A\int x^{-2}\,dx = -\frac{A}{x}.$$

Second term: for $$\displaystyle \int B\,dx$$, $$B$$ is a constant, hence

$$\int B\,dx = Bx.$$

Combining these results, the definite integral becomes

$$F = Gm\left[\,-\frac{A}{x} + Bx\;\right]_{x=a}^{x=a+L}.$$

We substitute the limits. First, the upper limit $$x = a + L$$:

Upper contribution  $$= -\frac{A}{a + L} + B(a + L).$$

Next, the lower limit $$x = a$$:

Lower contribution  $$= -\frac{A}{a} + Ba.$$

Subtracting the lower contribution from the upper gives the net value of the bracketed expression:

$$\begin{aligned} F &= Gm\Bigg[\left(-\frac{A}{a + L} + B(a + L)\right) \;-\; \left(-\frac{A}{a} + Ba\right)\Bigg] \\ &= Gm\Bigg[-\frac{A}{a + L} + B(a + L) + \frac{A}{a} - Ba\Bigg]. \end{aligned}$$

The two $$Ba$$ terms cancel:

$$B(a + L) - Ba \;=\; BL.$$

Hence the expression inside the brackets simplifies to

$$-\frac{A}{a + L} + \frac{A}{a} + BL.$$

Reordering the terms more conventionally, we write

$$\frac{A}{a} - \frac{A}{a + L} + BL.$$

Therefore the magnitude of the gravitational force is

$$F = Gm\left[A\!\left(\frac{1}{a} - \frac{1}{a + L}\right) + BL\right].$$

This matches the expression given in Option D.

Hence, the correct answer is Option D.

For a simple pendulum the time‐period is given by the well-known formula

$$T = 2\pi\sqrt{\dfrac{l}{g}}\; ,$$

where $$l$$ is the length of the pendulum and $$g$$ is the acceleration due to gravity at the place of observation.

On Earth the period is provided as $$T_E = 2\ \text{s}.$$ So, on the Earth we have

$$2 = 2\pi\sqrt{\dfrac{l}{g_E}} \; ,$$

which we shall keep for later comparison. The length $$l$$ of the pendulum is the same everywhere; only $$g$$ changes from one celestial body to another.

Now we consider the given planet. Its mass is three times that of Earth, so

$$M_P = 3M_E.$$

Its diameter is also three times that of Earth, therefore its radius is three times Earth’s radius:

$$R_P = 3R_E.$$

The surface acceleration due to gravity is determined by Newton’s universal-gravitation expression

$$g = \dfrac{GM}{R^{2}},$$

where $$G$$ is the gravitational constant, $$M$$ is the mass of the body, and $$R$$ is its radius.

Applying this to the planet we write

$$g_P = \dfrac{G\,M_P}{R_P^{2}}.$$

Substituting $$M_P = 3M_E$$ and $$R_P = 3R_E$$ gives

$$g_P = \dfrac{G\,(3M_E)}{(3R_E)^{2}} = \dfrac{3G M_E}{9R_E^{2}} = \dfrac{1}{3}\,\dfrac{G M_E}{R_E^{2}} = \dfrac{g_E}{3}.$$

Thus the planet’s gravitational acceleration is one-third that of Earth:

$$g_P = \dfrac{g_E}{3}.$$

We now compute the pendulum’s period on the planet. Using the same length $$l$$ in the original period formula, we have

$$T_P = 2\pi\sqrt{\dfrac{l}{g_P}} = 2\pi\sqrt{\dfrac{l}{g_E/3}} = 2\pi\sqrt{\dfrac{3l}{g_E}} = \sqrt{3}\,\bigl(2\pi\sqrt{\dfrac{l}{g_E}}\bigr).$$

The expression in large parentheses is exactly the Earth period $$T_E$$, so

$$T_P = \sqrt{3}\,T_E.$$

Given $$T_E = 2\ \text{s},$$ we substitute:

$$T_P = \sqrt{3}\times 2 = 2\sqrt{3}\ \text{s}.$$

Hence, the correct answer is Option D.

$$\frac{g_e}{g_p} = \frac{M_e}{M_p} \times \left(\frac{R_p}{R_e}\right)^2$$

$$\frac{9}{4} = 9 \times \left(\frac{R_p}{R}\right)^2 \implies \left(\frac{R_p}{R}\right)^2 = \frac{1}{4}$$

$$R_p = \frac{R}{2}$$

Let us denote the centre of both the solid sphere and the surrounding shell by $$O$$. The given data are:

• Solid sphere: mass $$M$$, radius $$a$$.
• Concentric spherical shell: inner radius $$a$$, outer radius $$(a+2a)=3a$$, mass $$2M$$, uniform mass distribution.

We want the gravitational field (its magnitude) at a point situated at a distance $$r=3a$$ from the centre, i.e. at the outer surface of the shell.

First, recall the basic result for spherically symmetric mass distributions:
Statement of the theorem — For any point lying outside a spherically symmetric mass distribution (or exactly on its outer surface), the gravitational field is the same as if the entire mass were concentrated at the centre.

Here the point of interest is on the outer surface, so the above theorem applies. Therefore, we may replace the complete two-part body by a single point mass at the centre whose value equals the total mass enclosed within radius $$3a$$.

Now compute that total mass:

$$M_{\text{total}} = \text{mass of solid sphere} + \text{mass of shell} = M + 2M = 3M.$$

Using Newton’s law for the magnitude of gravitational field $$g$$ produced by a point mass $$M_{\text{total}}$$ at a distance $$r$$,

$$g = \frac{G\,M_{\text{total}}}{r^{2}},$$

where $$G$$ is the universal gravitational constant.

Substituting $$M_{\text{total}} = 3M$$ and $$r = 3a$$ gives

$$g = \frac{G \,(3M)}{(3a)^{2}} = \frac{3GM}{9a^{2}} = \frac{GM}{3a^{2}}.$$

Hence, the correct answer is Option B.

We have a continuous spherically-symmetric mass distribution whose density varies with the radial distance as $$\rho(r)=\dfrac{K}{r^{2}}$$, where $$K$$ is a constant. To find the relation between the orbital radius $$R$$ and the period $$T$$ of a light test particle moving in a circular orbit, we first determine the mass enclosed within any radius $$r$$.

The formula for mass enclosed inside a sphere of radius $$r$$ for a density distribution $$\rho(r)$$ is

$$M(r)=\displaystyle\int_{0}^{r}4\pi r'^{2}\,\rho(r')\,dr'.$$

Substituting $$\rho(r')=\dfrac{K}{r'^{2}}$$ gives

$$M(r)=\int_{0}^{r}4\pi r'^{2}\left(\dfrac{K}{r'^{2}}\right)dr'$$ $$=4\pi K\int_{0}^{r}dr'$$ $$=4\pi K\,r.$$

Thus the total mass inside the orbit grows linearly with radius:

$$M(r)=4\pi K\,r.$$

Next, the magnitude of the gravitational field at distance $$r$$ (for a spherically symmetric mass distribution) is given by Newton’s law of gravitation,

$$g(r)=\dfrac{G\,M(r)}{r^{2}},$$

where $$G$$ is the gravitational constant. Substituting $$M(r)=4\pi K\,r$$ yields

$$g(r)=\dfrac{G\,(4\pi K\,r)}{r^{2}}=\dfrac{4\pi G K}{r}.$$

Now consider a particle of mass $$m$$ in a circular orbit of radius $$R$$ and speed $$v$$. The required centripetal force is supplied by gravity, so we equate

$$\dfrac{m v^{2}}{R}=m\,g(R)=m\left(\dfrac{4\pi G K}{R}\right).$$

Canceling the common factor $$m$$ and the radius $$R$$ on both sides, we get

$$v^{2}=4\pi G K.$$

Thus $$v$$ is a constant (independent of $$R$$):

$$v=\sqrt{4\pi G K}.$$

The time period $$T$$ of the circular orbit is related to the speed $$v$$ through the usual relation for one full revolution,

$$T=\dfrac{\text{circumference}}{\text{speed}}=\dfrac{2\pi R}{v}.$$

Substituting the constant value of $$v$$ obtained above, we have

$$T=\dfrac{2\pi R}{\sqrt{4\pi G K}}.$$

Since the denominator is a constant, this expression shows a direct proportionality:

$$\dfrac{T}{R}=\dfrac{2\pi}{\sqrt{4\pi G K}}=\text{constant}.$$

Hence $$T/R$$ remains the same for any circular orbit in this density field, while all other listed combinations vary with $$R$$. This matches option C.

Hence, the correct answer is Option C.

We begin by recalling the condition for uniform circular motion. In a circle of radius $$R$$, the required centripetal force is given by the well-known formula

$$F_{\text{centripetal}}=\frac{m v^{2}}{R},$$

where $$m$$ is the mass of the particle and $$v$$ is its constant linear speed along the orbit.

According to the statement of the problem, the only force acting is a central force that varies inversely as the $$n^{\text{th}}$$ power of the distance $$R$$. Hence we can write this central force as

$$F_{\text{central}}=\frac{k}{R^{\,n}},$$

where $$k$$ is a positive proportionality constant (its exact value is immaterial for the proportionality we seek).

For the particle to remain in uniform circular motion, the central force must supply exactly the needed centripetal force. Therefore we equate the two expressions:

$$\frac{m v^{2}}{R}=\frac{k}{R^{\,n}}.$$

Now we isolate $$v^{2}$$ by multiplying both sides by $$R$$:

$$m v^{2}=k\,R^{\,1-n}.$$

Next we divide by $$m$$ to solve for $$v^{2}$$:

$$v^{2}=\frac{k}{m}\,R^{\,1-n}.$$

Taking the square root of both sides gives the speed $$v$$ explicitly:

$$v=\sqrt{\frac{k}{m}}\;R^{\,\frac{1-n}{2}}.$$

The time period $$T$$ of one complete revolution is the circumference $$2\pi R$$ divided by the speed $$v$$. Stating this definition, we have

$$T=\frac{2\pi R}{v}.$$

Substituting the expression we have just obtained for $$v$$, we write

$$T=\frac{2\pi R}{\sqrt{\frac{k}{m}}\;R^{\,\frac{1-n}{2}}}.$$

To simplify, notice that dividing by a power of $$R$$ subtracts the exponents. First, let us handle the constant factors:

$$\frac{2\pi R}{\sqrt{\frac{k}{m}}}=2\pi\sqrt{\frac{m}{k}}\;R,$$

because $$\displaystyle\frac{1}{\sqrt{k/m}}=\sqrt{\frac{m}{k}}.$$

Now place the remaining power of $$R$$ in the denominator:

$$T = 2\pi\sqrt{\frac{m}{k}}\;R \;R^{-\frac{1-n}{2}}.$$

Using the law of indices $$R^{a}R^{b}=R^{\,a+b}$$, we combine the exponents of $$R$$:

$$T = 2\pi\sqrt{\frac{m}{k}}\;R^{\,1-\frac{1-n}{2}}.$$

Next we simplify the exponent. The exponent is

$$1-\frac{1-n}{2}=\frac{2}{2}-\frac{1-n}{2}=\frac{2-(1-n)}{2}.$$

Expanding the parentheses in the numerator gives

$$2-(1-n)=2-1+n=1+n.$$

Hence the exponent becomes $$\dfrac{1+n}{2}$$, and we obtain

$$T = 2\pi\sqrt{\frac{m}{k}}\;R^{\frac{\,1+n}{2}}.$$

The factor $$2\pi\sqrt{m/k}$$ is a constant for a given system, so the proportionality we seek is

$$T\;\propto\;R^{\frac{n+1}{2}}.$$

This matches Option D exactly. All other options give different powers of $$R$$, so they are incorrect.

Hence, the correct answer is Option 4.

For a body of mass $$m$$ moving around a planet of mass $$M$$ in a circular orbit of radius $$R$$, the total mechanical energy of the system is the sum of its kinetic and potential energies. For every circular orbit we use the well-known result

$$E \;=\; -\,\frac{G M m}{2 r},$$

where $$r$$ is the orbital radius and $$G$$ is the universal gravitational constant.

Initially the whole mass $$m$$ is in the orbit of radius $$R$$, so the initial total energy is

$$E_i \;=\; -\,\frac{G M m}{2R}.$$

At the given instant, the body splits into two equal parts, each of mass $$\dfrac{m}{2}$$. We neglect the gravitational interaction between the two fragments themselves and consider only their interaction with the planet.

The first fragment of mass $$\dfrac{m}{2}$$ now moves in a circular orbit of radius $$\dfrac{R}{2}$$. Using the same energy formula with $$r=\dfrac{R}{2}$$ we get

$$E_1 \;=\; -\,\frac{G M \left(\dfrac{m}{2}\right)}{2\left(\dfrac{R}{2}\right)} \;=\; -\,\frac{G M \left(\dfrac{m}{2}\right)}{R} \;=\; -\,\frac{G M m}{2R}.$$

The second fragment of mass $$\dfrac{m}{2}$$ moves in a circular orbit of radius $$\dfrac{3R}{2}$$. Inserting $$r=\dfrac{3R}{2}$$ gives

$$E_2 \;=\; -\,\frac{G M \left(\dfrac{m}{2}\right)}{2\left(\dfrac{3R}{2}\right)} \;=\; -\,\frac{G M \left(\dfrac{m}{2}\right)}{3R} \;=\; -\,\frac{G M m}{6R}.$$

Adding the energies of both fragments yields the final total energy of the system:

$$E_f \;=\; E_1 + E_2 \;=\; \left(-\,\frac{G M m}{2R}\right) + \left(-\,\frac{G M m}{6R}\right) \;=\; -\,\frac{3G M m}{6R} - \frac{G M m}{6R} \;=\; -\,\frac{4G M m}{6R} \;=\; -\,\frac{2G M m}{3R}.$$

Now we calculate the change in energy, i.e. the difference between the final and the initial energies:

$$\Delta E \;=\; E_f - E_i \;=\; \left(-\,\frac{2G M m}{3R}\right) - \left(-\,\frac{G M m}{2R}\right).$$

Combining the two terms by bringing them to a common denominator $$6R$$, we obtain

$$\Delta E \;=\; -\,\frac{4G M m}{6R} + \frac{3G M m}{6R} \;=\; -\,\frac{G M m}{6R}.$$

The change is negative, indicating that the total mechanical energy of the system decreases in magnitude by $$\dfrac{G M m}{6R}$$.

Hence, the correct answer is Option C.

We begin by recalling the definition of the true (gravitational) weight of a body of mass $$m$$ on the Earth, which is given by $$W_g = m g,$$ where $$g$$ is the magnitude of the acceleration due to gravity directed toward the centre of the Earth.

However, because the Earth is rotating with angular velocity $$\omega,$$ every point on its surface (except the poles) experiences an outward centrifugal acceleration. The formula for the centrifugal acceleration at any latitude is

$$a_c = \omega^2 r,$$

where $$r$$ is the perpendicular distance of the point from the axis of rotation. If the latitude of the point is denoted by $$\phi$$ (zero at the equator and $$90^{\circ}$$ at the poles), then the geometry of the Earth gives

$$r = R \cos\phi,$$

where $$R$$ is the radius of the Earth. Substituting this expression into the centrifugal-acceleration formula yields

$$a_c \;=\; \omega^2 \, (R\cos\phi).$$

The centrifugal acceleration is directed horizontally outward, but we need only the component that opposes gravity, i.e., the component along the radial line from the Earth’s centre. Geometry again tells us that this component is

$$a_{c,\text{radial}} = a_c \cos\phi = \omega^2 R \cos^2\phi.$$

The effective downward acceleration that actually presses the body against the ground (and hence decides the reading on a weighing machine) is therefore

$$g_{\text{eff}} = g \;-\; \omega^2 R \cos^2\phi.$$

Accordingly, the apparent or measured weight becomes

$$W_{\text{apparent}} = m g_{\text{eff}} \;=\; m\Bigl(g - \omega^2 R \cos^2\phi\Bigr).$$

Now imagine that the Earth’s angular velocity of rotation is increased from its present value to some larger value $$\omega'$$ with $$\omega' > \omega.$$ In the expression for $$g_{\text{eff}}$$ we must then replace $$\omega$$ by $$\omega'.$$ Because $$\omega'$$ is larger, the term $$\omega'^2 R\cos^2\phi$$ is also larger, while $$g$$ and $$R$$ remain essentially unchanged. Thus

$$g_{\text{eff,new}} = g - \omega'^2 R \cos^2\phi \;<\; g - \omega^2 R \cos^2\phi = g_{\text{eff,old}}$$

for every latitude where $$\cos\phi \neq 0,$$ that is, for every point except the poles.

At the poles themselves, the latitude is $$\phi = 90^{\circ},$$ so $$\cos\phi = 0.$$ In that special case the centrifugal term vanishes identically, and we have

$$g_{\text{eff,pole}} = g$$

both before and after the increase in $$\omega.$$ Hence there is no change in weight at the poles.

Collecting these observations, we see that increasing the Earth’s spin causes the apparent weight to decrease everywhere except at the two poles, where it remains the same.

Hence, the correct answer is Option 3.

We begin with Newton’s universal law of gravitation, which gives the magnitude of the force exerted by a body of mass $$M$$ on a small test mass $$m$$ situated at a distance $$r$$ as

$$F \;=\; \dfrac{G\,M\,m}{r^{2}},$$

where $$G$$ is the gravitational constant.

The earth is a sphere of radius $$R = 6400\;\text{km}$$. For a celestial body whose centre is at a distance $$D$$ from the centre of the earth, the distance of that body from

• the nearest point on the earth’s surface is $$D - R$$,

• the farthest point on the earth’s surface is $$D + R$$.

Hence the forces on a unit test mass at these two points are

$$F_{\text{near}} = \dfrac{G\,M}{(D - R)^{2}}, \qquad F_{\text{far}} = \dfrac{G\,M}{(D + R)^{2}}.$$

The difference (which is what produces the tidal effect) is therefore

$$\Delta F \;=\; F_{\text{near}} \;-\; F_{\text{far}} \;=\; G\,M \!\left[ \dfrac{1}{(D - R)^{2}} \;-\; \dfrac{1}{(D + R)^{2}} \right].$$

Because the earth’s radius is very small compared with any celestial distance (i.e. $$R \ll D$$), we expand each denominator by writing

$$(D \pm R)^{-2} \;=\; D^{-2}\,(1 \pm \tfrac{R}{D})^{-2} \;\approx\; D^{-2}\,\bigl(1 \mp 2\tfrac{R}{D}\bigr),$$

using the binomial approximation $$(1 + x)^{n} \approx 1 + nx$$ for small $$x$$.

Substituting the expanded forms, we get

$$\dfrac{1}{(D - R)^{2}} \approx D^{-2}\!\left(1 + 2\frac{R}{D}\right), \qquad \dfrac{1}{(D + R)^{2}} \approx D^{-2}\!\left(1 - 2\frac{R}{D}\right).$$

Taking their difference yields

$$\dfrac{1}{(D - R)^{2}} - \dfrac{1}{(D + R)^{2}} \approx D^{-2}\!\Bigl(1 + 2\frac{R}{D} - 1 + 2\frac{R}{D}\Bigr) \;=\; 4\,\dfrac{R}{D^{3}}.$$

Hence

$$\boxed{\;\Delta F \approx 4\,G\,M\,\dfrac{R}{D^{3}}\;}.$$

Notice that $$G$$, the earth’s radius $$R$$ and the test mass have the same values for both the moon and the sun, so they will cancel when we form the required ratio. Writing

$$\Delta F_{1} \text{ (moon)} = 4\,G\,M_{\text{moon}}\dfrac{R}{D_{\text{moon}}^{\,3}},$$ $$\Delta F_{2} \text{ (sun)} = 4\,G\,M_{\text{sun }}\dfrac{R}{D_{\text{sun }}^{\,3}},$$

we obtain

$$\dfrac{\Delta F_{1}}{\Delta F_{2}} = \dfrac{M_{\text{moon}}}{M_{\text{sun}}} \left(\dfrac{D_{\text{sun}}}{D_{\text{moon}}}\right)^{\!3}.$$

Now we substitute the numerical data (the units are all consistent because every distance is in kilometres and every mass in kilograms):

$$M_{\text{moon}} = 8 \times 10^{22}, \qquad M_{\text{sun}} = 2 \times 10^{30},$$

$$D_{\text{moon}} = 0.4 \times 10^{6}, \qquad D_{\text{sun}} = 150 \times 10^{6}.$$

First the mass ratio:

$$\dfrac{M_{\text{moon}}}{M_{\text{sun}}} = \dfrac{8 \times 10^{22}}{2 \times 10^{30}} = 4 \times 10^{-8}.$$

Next the distance ratio:

$$\dfrac{D_{\text{sun}}}{D_{\text{moon}}} = \dfrac{150 \times 10^{6}}{0.4 \times 10^{6}} = \dfrac{150}{0.4} = 375.$$

Its cube is

$$375^{3} = (375)^{2}\times 375 = 140\,625 \times 375 = 52\,734\,375.$$

Finally, combining everything,

$$\dfrac{\Delta F_{1}}{\Delta F_{2}} = \bigl(4 \times 10^{-8}\bigr)\; \times 52\,734\,375 = 2.109375 \; \approx \; 2.$$

Thus the number closest to $$\dfrac{\Delta F_{1}}{\Delta F_{2}}$$ is $$2$$.

Hence, the correct answer is Option A.

We are given a spherical body with mass density $$\rho(r) = \frac{k}{r}$$ for $$r \le R$$ and $$\rho(r) = 0$$ for $$r > R$$. We need to find the acceleration $$a$$ of a test particle as a function of $$r$$.

Step 1: Find the mass enclosed within radius $$r$$ (for $$r \le R$$).

Using the mass element in a spherical shell of thickness $$dr$$:

$$dM = \rho(r) \cdot 4\pi r^2 \, dr = \frac{k}{r} \cdot 4\pi r^2 \, dr = 4\pi k \, r \, dr$$

Integrating from 0 to $$r$$:

$$M(r) = \int_0^r 4\pi k \, r' \, dr' = 4\pi k \cdot \frac{r^2}{2} = 2\pi k r^2$$

Step 2: Find the gravitational acceleration for $$r \le R$$.

By Newton's law of gravitation (using the shell theorem, only mass enclosed within radius $$r$$ contributes):

$$a(r) = \frac{G \cdot M(r)}{r^2} = \frac{G \cdot 2\pi k r^2}{r^2} = 2\pi G k$$

This is a constant, independent of $$r$$. So for $$r \le R$$, the acceleration is constant.

Step 3: Find the gravitational acceleration for $$r > R$$.

The total mass of the sphere is $$M(R) = 2\pi k R^2$$.

For $$r > R$$, the entire sphere acts as a point mass at the center:

$$a(r) = \frac{G \cdot M(R)}{r^2} = \frac{2\pi G k R^2}{r^2}$$

This decreases as $$\frac{1}{r^2}$$.

Step 4: Describe the graph.

The acceleration $$a$$ is constant for $$r \le R$$, and then drops as $$\frac{1}{r^2}$$ for $$r > R$$.

The correct answer is Option C.

1. Inside the earth ($$d < R$$):

$$g_{in} = \frac{G \cdot M_{in}}{d^2}$$

$$M_{in} = M \left( \frac{d^3}{R^3} \right)$$

$$g_{in} = \frac{G \cdot M \cdot d^3}{R^3 \cdot d^2} = \left( \frac{GM}{R^3} \right) d$$

2. Outside earth's surface ($$d >R$$)

$$g_{out} = \frac{GM}{d^2}$$

Graph:

From $$d = 0$$ to $$d = R$$, the graph must be a linearly increasing straight line.

From $$d = R$$ onwards, the graph must be a decreasing curve that asymptotically approaches the horizontal axis.

We begin by recalling the two standard formulae that govern motion under gravity around the earth:

1. The magnitude of the gravitational acceleration at the earth’s surface is related to the earth’s mass by $$g \;=\;\dfrac{G\,M}{R^{2}}.$$

2. For any circular orbit of radius $$r$$ about the earth, the orbital speed $$v_\text{orbit}$$ is given by $$v_\text{orbit}\;=\;\sqrt{\dfrac{G\,M}{r}}.$$

3. From the same radius $$r$$, the escape speed $$v_\text{escape}$$ is obtained from the energy condition $$\dfrac{1}{2}m v_\text{escape}^{2}\;=\;\dfrac{G\,M\,m}{r},$$ whence $$v_\text{escape}\;=\;\sqrt{\dfrac{2\,G\,M}{r}}.$$

The satellite in the problem is already moving in a circular orbit of radius

$$r \;=\; R + h,$$

where $$R$$ is the radius of the earth and $$h$$ is the altitude of the satellite. The given condition $$h \ll R$$ will allow a useful approximation later.

Using formula 2, the present orbital speed of the satellite is

$$v_\text{orbit}\;=\;\sqrt{\dfrac{G\,M}{R+h}}.$$

Using formula 3, the escape speed needed from the same point is

$$v_\text{escape}\;=\;\sqrt{\dfrac{2\,G\,M}{R+h}}.$$

The satellite must increase its speed by the difference of these two values. Hence the required increment $$\Delta v$$ is

$$\Delta v \;=\; v_\text{escape} \;-\; v_\text{orbit}$$ $$=\;\sqrt{\dfrac{2\,G\,M}{R+h}} \;-\;\sqrt{\dfrac{G\,M}{R+h}}.$$

We factor out the common square-root:

$$\Delta v \;=\;\sqrt{\dfrac{G\,M}{R+h}}\;\Bigl(\sqrt{2}\;-\;1\Bigr).$$

Next we eliminate $$G\,M$$ in favour of the surface quantity $$g$$ by substituting $$G\,M = g\,R^{2}$$ from formula 1:

$$\sqrt{\dfrac{G\,M}{R+h}} \;=\; \sqrt{\dfrac{g\,R^{2}}{R+h}}.$$

Because $$h \ll R,$$ we may approximate $$R+h \approx R.$$ This gives

$$\sqrt{\dfrac{g\,R^{2}}{R+h}} \;\approx\; \sqrt{\dfrac{g\,R^{2}}{R}} \;=\; \sqrt{g\,R}.$$

Substituting this back, we obtain

$$\Delta v \;\approx\; \sqrt{g\,R}\,\Bigl(\sqrt{2}\;-\;1\Bigr).$$

This expression matches Option B.

Hence, the correct answer is Option B.

Kepler's Second Law states that a line joining a planet and the Sun sweeps out equal areas during equal intervals of time.

$$t \propto \text{Area swept}$$

The minor axis $$CA$$ divides the ellipse into two equal halves. Therefore, the area of the elliptical sector to the left of line $$CA$$ is exactly $$\frac{A}{2}$$.

$$\text{Area swept over path } CDA = (\text{Area of the half-ellipse to the left of } CA) - (\text{Area of } \triangle CSA)$$

$$\text{Area swept for } t_2 = \frac{A}{2} - \frac{A}{4} = \frac{A}{4}$$

$$\text{Area swept for } t_1 = \text{Total Area} - (\text{Area swept for } t_2)$$

$$\text{Area swept for } t_1 = A - \frac{A}{4} = \frac{3A}{4}$$

$$\frac{t_1}{t_2} = \frac{\text{Area for } t_1}{\text{Area for } t_2}$$

$$\frac{t_1}{t_2} = \frac{3A/4}{A/4} = 3$$

$$t_1 = 3t_2$$

We start by recalling Newton’s universal law of gravitation, which states that the attractive force between two point masses is

$$F=\dfrac{G\,m_1\,m_2}{r^{\,2}},$$

where $$G$$ is the universal gravitational constant, $$m_1$$ and $$m_2$$ are the two masses involved, and $$r$$ is the separation between their centres.

In the present situation the two masses are the earth of mass $$M$$ and the astronaut of mass $$m$$. The distance between their centres is not the earth’s radius $$R$$ alone, because the satellite is orbiting at a height $$h$$ above the earth’s surface. Hence the centre-to-centre distance becomes

$$r = R + h.$$

Substituting $$m_1 = M$$, $$m_2 = m$$, and $$r = R+h$$ into the gravitational formula, we obtain

$$F_G = \dfrac{G\,M\,m}{(R+h)^{2}}.$$

This expression gives the exact magnitude of the gravitational pull on the astronaut. It is clearly a positive, non-zero quantity as long as $$h$$ is finite. The sense of weightlessness experienced in orbit arises because both the satellite and the astronaut are in free fall with the same acceleration, not because the gravitational force vanishes.

Among the given alternatives, Option C reproduces precisely the derived formula, while the remaining choices either set the force to zero or express an inequality instead of the exact value.

Hence, the correct answer is Option C.

Consider a very long cylindrical galaxy of length $$L$$ and radius $$R$$ (with $$R \ll L$$), with mass uniformly distributed. A star orbits the galaxy in a plane perpendicular to the galaxy's axis and passing through its center, at a distance $$r$$ from the axis (where $$r > R$$ since the star is outside the galaxy). We need to find the relationship between the orbital time period $$T$$ and $$r$$.

Due to the cylindrical symmetry and the condition $$R \ll L$$, we can approximate the galaxy as an infinitely long cylinder for gravitational effects at distances $$r$$ (where $$r \ll L$$). The mass per unit length $$\lambda$$ is constant and given by the total mass divided by $$L$$. For an infinitely long cylinder, the gravitational field outside the cylinder depends only on $$\lambda$$.

Using Gauss's law for gravity, consider a cylindrical Gaussian surface of radius $$r$$ and length $$l$$, coaxial with the galaxy. The mass enclosed by this surface is $$\lambda l$$. The gravitational flux through the curved surface is $$g \cdot 2\pi r l$$ (since the field is radial and uniform), and through the ends is zero (as the field is parallel to the ends). Gauss's law states:

$$\int \mathbf{g} \cdot d\mathbf{A} = -4\pi G M_{\text{enc}}$$

Substituting the values:

$$g \cdot 2\pi r l = -4\pi G (\lambda l)$$

Solving for the magnitude of the gravitational field $$g$$ (ignoring the negative sign for direction):

$$g = \frac{2G\lambda}{r}$$

This gravitational force provides the centripetal force for the star's circular orbit. Let $$m$$ be the mass of the star and $$v$$ its orbital speed. The centripetal force is:

$$F_c = \frac{m v^2}{r}$$

The gravitational force is:

$$F_g = m g = m \cdot \frac{2G\lambda}{r}$$

Setting $$F_g = F_c$$:

$$m \cdot \frac{2G\lambda}{r} = \frac{m v^2}{r}$$

Canceling $$m$$ and $$r$$ (assuming $$m \neq 0$$ and $$r \neq 0$$):

$$\frac{2G\lambda}{r} = \frac{v^2}{r}$$

Multiplying both sides by $$r$$:

$$2G\lambda = v^2$$

Thus:

$$v^2 = 2G\lambda$$

Since $$\lambda$$ is constant for the galaxy, $$v$$ is constant and independent of $$r$$.

The orbital period $$T$$ is the time for one complete orbit, which is the circumference divided by the speed:

$$T = \frac{2\pi r}{v}$$

As $$v$$ is constant, $$T$$ is proportional to $$r$$:

$$T \propto r$$

Comparing with the options:

A. $$T \propto \sqrt{r}$$

B. $$T \propto r$$

C. $$T \propto r^2$$

D. $$T^2 \propto r^3$$

Option B matches the derived relationship.

Hence, the correct answer is Option B.

$$V_{\text{resultant}} = V_{\text{original sphere}} - V_{\text{removed sphere}}$$

The mass of the removed part is $$m = M \times \left( \frac{R/2}{R} \right)^3 = \frac{M}{8}$$

$$V_{\text{in}} = -\frac{GM}{2R^3} (3R^2 - x^2)$$

Substituting $$x = \frac{R}{2}$$:

$$V_{\text{orig}} = -\frac{GM}{2R^3} \left[ 3R^2 - \left(\frac{R}{2}\right)^2 \right] = -\frac{GM}{2R^3} \left( 3R^2 - \frac{R^2}{4} \right)$$

$$V_{\text{orig}} = -\frac{GM}{2R^3} \left( \frac{11R^2}{4} \right) = -\frac{11GM}{8R}$$

$$V_{\text{center}} = -\frac{3Gm}{2r}$$

Substituting $$m = \frac{M}{8}$$ and $$r = \frac{R}{2}$$:

$$V_{\text{rem}} = -\frac{3G(M/8)}{2(R/2)} = -\frac{3GM/8}{R} = -\frac{3GM}{8R}$$

$$V_P = V_{\text{orig}} - V_{\text{rem}}$$

$$V_P = \left( -\frac{11GM}{8R} \right) - \left( -\frac{3GM}{8R} \right)$$

$$V_P = -\frac{GM}{R}$$

For any point outside or on the surface, a spherical mass behaves as if all its mass $$M$$ is concentrated at its center.

$$V_{out} = -\frac{GM}{r}$$

This represents a rectangular hyperbola in the negative quadrant. As $$r \rightarrow \infty, V \rightarrow 0$$. At the surface ($$r = R$$), $$V_s = -\frac{GM}{R}$$.

For a uniform solid sphere, the potential at an internal point is given by:

$$V_{in} = -\frac{GM}{2R^3}(3R^2 - r^2)$$

This is a parabolic relation with respect to $$r$$. At the center ($$r = 0$$): $$V_c = -\frac{3GM}{2R} = 1.5 V_s$$. The potential at the center is $$1.5$$ times the potential at the surface (and more negative).

All these are correctly shown in option (c).

The mass of the removed sphere ($$m$$) is $$m = M \times \left( \frac{R/2}{R} \right)^3 = \frac{M}{8}$$

Center of the Original Sphere ($$O$$): $$(0,0)$$

Center of the Cavity ($$C$$): Since the cavity is carved between the center and periphery, its center is at $$(R/2, 0)$$.

Center of the External removed sphere ($$S$$): The distance between the center of the original sphere and the removed sphere is $$3R$$. Its position is $$(3R, 0)$$.

Distance from $$O$$ to $$S$$ ($$r_1$$) = $$3R$$. Distance from $$C$$ to $$S$$ ($$r_2$$) = $$3R - R/2 = 2.5R = \frac{5R}{2}$$

$$F_{net} = F_{original} - F_{cavity}$$

$$F_{net} = G \frac{M \cdot m}{r_1^2} - G \frac{m \cdot m}{r_2^2}$$

$$F_{net} = G \frac{M(M/8)}{(3R)^2} - G \frac{(M/8)(M/8)}{(5R/2)^2}$$

$$F_{net} = \frac{GM^2}{8 \cdot 9R^2} - \frac{GM^2}{64 \cdot \frac{25R^2}{4}}$$

$$F_{net} = \frac{GM^2}{72R^2} - \frac{GM^2}{16 \cdot 25R^2} = \frac{GM^2}{72R^2} - \frac{GM^2}{400R^2}$$

$$F_{net} = \frac{50GM^2}{3600R^2} - \frac{9GM^2}{3600R^2} = \frac{41GM^2}{3600R^2}$$

To estimate the time taken for Mangalyan to travel from Earth to Mars using a transfer orbit around the sun, we apply Kepler's laws. The transfer orbit is elliptical, with the sun at one focus. Mangalyan leaves Earth at perihelion (closest point to the sun) and meets Mars at aphelion (farthest point). The semi-major axis of Earth's orbit is given as $$a_e = 1.5 \times 10^{11}$$ m, and for Mars, $$a_m = 2.28 \times 10^{11}$$ m.

For the transfer orbit, the perihelion distance equals Earth's orbital radius ($$a_e$$), and the aphelion distance equals Mars' orbital radius ($$a_m$$). The semi-major axis of the transfer orbit ($$a_t$$) is the average of these distances:

$$ a_t = \frac{a_e + a_m}{2} = \frac{1.5 \times 10^{11} + 2.28 \times 10^{11}}{2} = \frac{3.78 \times 10^{11}}{2} = 1.89 \times 10^{11} \text{ m} $$

Kepler's third law states that the square of the orbital period ($$T$$) is proportional to the cube of the semi-major axis ($$a$$): $$T^2 \propto a^3$$. For Earth, the orbital period $$T_e$$ is 1 year (365 days), and the semi-major axis is $$a_e$$. For the transfer orbit, the period $$T_t$$ relates to $$a_t$$ by the same proportionality constant. Therefore, we can write:

$$ \frac{T_t^2}{T_e^2} = \frac{a_t^3}{a_e^3} $$

Solving for $$T_t$$:

$$ T_t = T_e \left( \frac{a_t}{a_e} \right)^{3/2} $$

The travel time from Earth to Mars is half the period of the transfer orbit, as Mangalyan travels from perihelion to aphelion, covering half the ellipse. Thus, the travel time $$\tau$$ is:

$$ \tau = \frac{T_t}{2} = \frac{T_e}{2} \left( \frac{a_t}{a_e} \right)^{3/2} $$

Substitute the values $$T_e = 365$$ days, $$a_t = 1.89 \times 10^{11}$$ m, and $$a_e = 1.5 \times 10^{11}$$ m:

$$ \frac{a_t}{a_e} = \frac{1.89 \times 10^{11}}{1.5 \times 10^{11}} = \frac{1.89}{1.5} = 1.26 $$

Now compute $$\left( \frac{a_t}{a_e} \right)^{3/2} = (1.26)^{3/2}$$:

First, find $$(1.26)^{3}$$:

$$ 1.26 \times 1.26 = 1.5876 $$

$$ 1.5876 \times 1.26 = 1.5876 \times 1.2 + 1.5876 \times 0.06 = 1.90512 + 0.095256 = 2.000376 $$

Now, $$(1.26)^{3/2} = \sqrt{(1.26)^3} = \sqrt{2.000376}$$. Using $$\sqrt{2} \approx 1.414213562$$ and approximating $$\sqrt{2.000376} \approx \sqrt{2} \times \sqrt{1.000188} \approx 1.414213562 \times (1 + \frac{0.000188}{2}) = 1.414213562 \times 1.000094 \approx 1.414213562 + 1.414213562 \times 0.000094$$:

$$ 1.414213562 \times 0.000094 = 0.000132936074828 $$

$$ \sqrt{2.000376} \approx 1.414213562 + 0.000132936074828 = 1.414346498074828 \approx 1.4143465 $$

Now substitute into the formula for $$\tau$$:

$$ \tau = \frac{365}{2} \times 1.4143465 = 182.5 \times 1.4143465 $$

Compute step by step:

$$ 182.5 \times 1.4 = 255.5 $$

$$ 182.5 \times 0.014 = 2.555 $$

$$ 182.5 \times 0.0003465 \approx 182.5 \times 0.0003 = 0.05475 \quad \text{and} \quad 182.5 \times 0.0000465 \approx 0.00848625 $$

$$ 0.0003465 \approx 0.0003 + 0.0000465 \quad \text{so} \quad 0.05475 + 0.00848625 = 0.06323625 $$

Sum the parts:

$$ 255.5 + 2.555 = 258.055 $$

$$ 258.055 + 0.06323625 = 258.11823625 \approx 258.12 \text{ days} $$

The calculated travel time is approximately 258.12 days. Comparing with the options:

A. 220 days

B. 500 days

C. 260 days

D. 320 days

The value 258.12 days is closest to 260 days.

Hence, the correct answer is Option C.

We begin by fixing our picture. The four identical particles, each of mass $$M$$, are placed at the four vertices of a square that is inscribed in a circle of radius $$R$$. The centre of the circle is also the centre of the square. Because the particles are equally spaced (every $$90^{\circ}$$), the arrangement is

A: $$\left(R,0\right)$$, B: $$\left(0,R\right)$$, C: $$\left(-R,0\right)$$, D: $$\left(0,-R\right)$$.

The side length of the square is obtained from the well-known relation “diagonal of a square $$= \sqrt{2}\,$$ (side)”. Here the diagonal $$AC$$ equals the diameter $$2R$$, so

$$\text{side}\;s=\frac{2R}{\sqrt{2}}=\sqrt{2}\,R.$$

Thus the distance between adjacent particles is $$s=\sqrt{2}\,R$$ and the distance between opposite particles is the diameter $$2R$$.

We now calculate, one by one, the gravitational forces on particle $$A$$ due to the other three particles.

1. Magnitude of the force exerted by each adjacent particle (for example, by $$B$$ or $$D$$): we use Newton’s law of gravitation,

$$F=\frac{G\,M\,M}{r^{2}},$$

and insert $$r=s=\sqrt{2}\,R$$, giving

$$F_{\text{adj}}=\frac{G M^{2}}{\left(\sqrt{2}\,R\right)^{2}} =\frac{G M^{2}}{2R^{2}}.$$

2. Magnitude of the force exerted by the opposite particle $$C$$ (distance $$2R$$):

$$F_{\text{opp}}=\frac{G M^{2}}{(2R)^{2}} =\frac{G M^{2}}{4R^{2}}.$$

Next we resolve these vector forces into radial (towards the centre) components. Choosing the positive $$x$$-axis to point from the centre to particle $$A$$, the radial direction for $$A$$ is along the negative $$x$$-axis.

• Force on $$A$$ due to $$B$$. The displacement vector from $$A(R,0)$$ to $$B(0,R)$$ is $$(-R,\,R)$$. Its length is $$s=\sqrt{2}\,R$$, so the unit vector is $$\left(-\tfrac{1}{\sqrt{2}},\,\tfrac{1}{\sqrt{2}}\right).$$ Multiplying by the magnitude $$F_{\text{adj}}$$ we obtain

$$\vec F_{AB}=F_{\text{adj}} \left(-\frac{1}{\sqrt{2}},\,\frac{1}{\sqrt{2}}\right).$$

Its radial (negative $$x$$) component is therefore

$$F_{AB,x}=-\frac{F_{\text{adj}}}{\sqrt{2}}.$$

• Force on $$A$$ due to $$D$$ has a displacement vector $$(-R,-R)$$ and hence

$$F_{AD,x}=-\frac{F_{\text{adj}}}{\sqrt{2}}.$$

Adding the two adjacent contributions, the transverse $$y$$-components cancel and the combined radial contribution becomes

$$F_{\text{adj,\,rad}}=F_{AB,x}+F_{AD,x} =-\,\frac{F_{\text{adj}}}{\sqrt{2}} -\,\frac{F_{\text{adj}}}{\sqrt{2}} =-\,\sqrt{2}\,F_{\text{adj}}.$$

• Force on $$A$$ due to the opposite particle $$C$$ acts exactly along the diameter toward the centre, so its entire magnitude contributes radially:

$$F_{\text{opp,\,rad}}=-\,F_{\text{opp}}.$$

Hence the total radial (inward) gravitational force on particle $$A$$ is the algebraic sum of the two inward negatives, i.e. an inward magnitude

$$F_{\text{net}} =\sqrt{2}\,F_{\text{adj}}+F_{\text{opp}} =\sqrt{2}\left(\frac{G M^{2}}{2R^{2}}\right) +\left(\frac{G M^{2}}{4R^{2}}\right).$$

Simplifying term by term, first multiply out the coefficient of the adjacent term:

$$\sqrt{2}\,F_{\text{adj}} =\sqrt{2}\left(\frac{G M^{2}}{2R^{2}}\right) =\frac{\sqrt{2}\,G M^{2}}{2R^{2}} =\frac{2\sqrt{2}\,G M^{2}}{4R^{2}}.$$

Now place both terms over the common denominator $$4R^{2}$$:

$$F_{\text{net}} =\frac{2\sqrt{2}\,G M^{2}}{4R^{2}} +\frac{G M^{2}}{4R^{2}} =\frac{G M^{2}}{4R^{2}} \left(1+2\sqrt{2}\right).$$

This net gravitational attraction towards the centre must supply the necessary centripetal force for uniform circular motion. The centripetal force law is

$$\frac{M v^{2}}{R}=F_{\text{net}}.$$

Substituting the expression for $$F_{\text{net}}$$ gives

$$\frac{M v^{2}}{R} =\frac{G M^{2}}{4R^{2}}\left(1+2\sqrt{2}\right).$$

We now solve for $$v^{2}$$ by multiplying both sides by $$R/M$$:

$$v^{2} =\frac{G M}{4R}\left(1+2\sqrt{2}\right).$$

Taking the positive square root (speed is positive) we obtain the required speed of each particle:

$$v =\frac{1}{2}\sqrt{\frac{G M}{R}\left(1+2\sqrt{2}\right)}.$$

This expression exactly matches Option D.

Hence, the correct answer is Option D.

The gravitational field is given by $$\vec{g} = (5\hat{i} + 12\hat{j})$$ N kg$$^{-1}$$. We need to find the change in gravitational potential energy ($$\Delta U$$) for a particle of mass $$m = 2$$ kg when it is moved from the origin $$(0, 0)$$ to the point $$(7 \text{ m}, -3 \text{ m})$$.

The change in gravitational potential energy is given by $$\Delta U = -m \int_{\text{initial}}^{\text{final}} \vec{g} \cdot d\vec{r}$$. Since $$\vec{g}$$ is constant, the integral simplifies to a path-independent calculation.

Let $$d\vec{r} = dx\,\hat{i} + dy\,\hat{j}$$. Then, $$\vec{g} \cdot d\vec{r} = (5\hat{i} + 12\hat{j}) \cdot (dx\,\hat{i} + dy\,\hat{j}) = 5\,dx + 12\,dy$$.

So, the integral becomes:

$$$ \int_{(0,0)}^{(7,-3)} \vec{g} \cdot d\vec{r} = \int_{(0,0)}^{(7,-3)} (5\,dx + 12\,dy) $$$

We can compute this by integrating along the x-axis first and then the y-axis. From $$(0,0)$$ to $$(7,0)$$, $$y = 0$$ (so $$dy = 0$$), and then from $$(7,0)$$ to $$(7,-3)$$, $$x = 7$$ (so $$dx = 0$$).

First segment: from $$(0,0)$$ to $$(7,0)$$:

$$$ \int_{x=0}^{7} 5\,dx + \int_{y=0}^{0} 12\,dy = \int_{0}^{7} 5\,dx + 0 = 5 \left[ x \right]_{0}^{7} = 5 \times (7 - 0) = 35 $$$

Second segment: from $$(7,0)$$ to $$(7,-3)$$:

$$$ \int_{x=7}^{7} 5\,dx + \int_{y=0}^{-3} 12\,dy = 0 + \int_{0}^{-3} 12\,dy = 12 \left[ y \right]_{0}^{-3} = 12 \times (-3 - 0) = 12 \times (-3) = -36 $$$

Adding both segments:

$$$ \int_{(0,0)}^{(7,-3)} \vec{g} \cdot d\vec{r} = 35 + (-36) = -1 \text{ J kg}^{-1} $$$

Alternatively, using a straight-line path parameterization: let $$x = 7t$$, $$y = -3t$$, where $$t$$ goes from 0 to 1. Then $$dx = 7\,dt$$, $$dy = -3\,dt$$.

$$$ \vec{g} \cdot d\vec{r} = 5(7\,dt) + 12(-3\,dt) = 35\,dt - 36\,dt = -dt $$$

$$$ \int_{0}^{1} (-1)\,dt = -1 \left[ t \right]_{0}^{1} = -1 \times (1 - 0) = -1 \text{ J kg}^{-1} $$$

Both methods give the same result: $$-1$$ J kg$$^{-1}$$.

Now, the change in potential energy is:

$$$ \Delta U = -m \times \left( \int \vec{g} \cdot d\vec{r} \right) = - (2 \text{ kg}) \times (-1 \text{ J kg}^{-1}) = -2 \times (-1) = 2 \text{ J} $$$

Hence, the change in gravitational potential energy is 2 J.

Comparing with the options:

A. 71 J

B. $$13\sqrt{58}$$ J

C. 2 J

D. 1 J

So, the correct answer is Option C.

Since no external force acts on the two-planet system, the net linear momentum is conserved. Initially, both planets are at rest at infinite separation, so the total initial momentum is zero.

From conservation of linear momentum, $$m_1 v_1 - m_2 v_2 = 0$$, $$v_2 = \frac{m_1 v_1}{m_2} \quad \dots(1)$$

As gravity is a conservative force, the total mechanical energy of the system remains constant.

Total initial energy at $$r = \infty$$ is $$E_i = 0$$.

At separation $$d$$, the total energy is $$E_f = K_f + U_f = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2 - \frac{Gm_1 m_2}{d}$$

Equating $$E_i = E_f$$: $$\frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2 = \frac{Gm_1 m_2}{d}$$

Substituting the value of $$v_2$$ from equation (1):

$$\frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 \left( \frac{m_1 v_1}{m_2} \right)^2 = \frac{Gm_1 m_2}{d}$$

$$\frac{1}{2}m_1 v_1^2 \left( 1 + \frac{m_1}{m_2} \right) = \frac{Gm_1 m_2}{d}$$

$$\frac{1}{2}m_1 v_1^2 \left( \frac{m_1 + m_2}{m_2} \right) = \frac{Gm_1 m_2}{d}$$

$$v_1^2 = \frac{2Gm_2^2}{d(m_1 + m_2)}$$

$$v_1 = m_2 \sqrt{\frac{2G}{d(m_1 + m_2)}}$$

By symmetry or by substituting $$v_1$$ back into equation (1):

$$v_2 = m_1 \sqrt{\frac{2G}{d(m_1 + m_2)}}$$

The gravitational field is given by $$\vec{E} = (5 \text{N/kg}) \hat{i} + (12 \text{N/kg}) \hat{j}$$. The potential at the origin (0,0) is zero. We need to find the ratio of the potential at point (12 m, 0) to the potential at point (0, 5 m).

The gravitational potential $$V$$ is related to the gravitational field $$\vec{E}$$ by $$\vec{E} = -\nabla V$$. Therefore, the potential difference between two points is given by the line integral:

$$$ V_B - V_A = -\int_A^B \vec{E} \cdot d\vec{r} $$$

Since the potential at the origin is zero, we can compute the potential at any point by integrating from (0,0) to that point. The field is conservative, so the path of integration does not matter. We will use straight-line paths for simplicity.

First, calculate the potential at point P(12 m, 0). We integrate along the x-axis from (0,0) to (12,0). Along this path, $$y = 0$$ and $$dy = 0$$, so $$d\vec{r} = dx \hat{i}$$. The dot product is:

$$$ \vec{E} \cdot d\vec{r} = (5 \hat{i} + 12 \hat{j}) \cdot (dx \hat{i}) = 5 dx $$$

The potential at P is:

$$$ V_P - V_{\text{origin}} = -\int_{(0,0)}^{(12,0)} \vec{E} \cdot d\vec{r} = -\int_{0}^{12} 5 dx $$$

Solving the integral:

$$$ -\int_{0}^{12} 5 dx = -5 \int_{0}^{12} dx = -5 \left[ x \right]_{0}^{12} = -5 (12 - 0) = -60 $$$

Since $$V_{\text{origin}} = 0$$, we have $$V_P = -60 \text{J/kg}$$.

Next, calculate the potential at point Q(0, 5 m). We integrate along the y-axis from (0,0) to (0,5). Along this path, $$x = 0$$ and $$dx = 0$$, so $$d\vec{r} = dy \hat{j}$$. The dot product is:

$$$ \vec{E} \cdot d\vec{r} = (5 \hat{i} + 12 \hat{j}) \cdot (dy \hat{j}) = 12 dy $$$

The potential at Q is:

$$$ V_Q - V_{\text{origin}} = -\int_{(0,0)}^{(0,5)} \vec{E} \cdot d\vec{r} = -\int_{0}^{5} 12 dy $$$

Solving the integral:

$$$ -\int_{0}^{5} 12 dy = -12 \int_{0}^{5} dy = -12 \left[ y \right]_{0}^{5} = -12 (5 - 0) = -60 $$$

Since $$V_{\text{origin}} = 0$$, we have $$V_Q = -60 \text{J/kg}$$.

The ratio of the potentials is:

$$$ \frac{V_P}{V_Q} = \frac{-60}{-60} = 1 $$$

Hence, the ratio is 1, which corresponds to Option B.

So, the answer is Option B.

Mass of removed part, $$m = \frac{\text{Volume of removed part}}{\text{Volume of whole sphere}} \times M = \frac{\frac{4}{3}\pi (R/2)^3}{\frac{4}{3}\pi R^3} \times M = \frac{M}{8}$$

The gravitational field $$E$$ at a distance $$x$$ from a point mass (or a uniform sphere at a far point) is $$E = \frac{GM}{x^2}$$.

Since $$P$$ is at a distance $$x$$ from the center, $$E_{total} = \frac{GM}{x^2}$$.

$$E_{removed} \approx \frac{G(M/8)}{x^2}$$

$$E_{net} = E_{total} - E_{removed}$$

$$E_{net} \approx \frac{GM}{x^2} - \frac{GM}{8x^2}$$

$$E_{net} \approx \left(1 - \frac{1}{8}\right) \frac{GM}{x^2} = \mathbf{\frac{7}{8} \frac{GM}{x^2}}$$

Position 1: Solar Eclipse

The Moon is located between the Earth and the Sun. In this case, both the Sun and the Moon pull the Earth in the same direction (towards the Sun).

$$a_1 = a_s + a_m$$

Position 2: Other side (Full Moon)

The Moon is on the opposite side of the Earth, away from the Sun. The Sun pulls the Earth towards itself, while the Moon pulls the Earth in the opposite direction.

$$a_2 = a_s - a_m$$

The change in the value of acceleration ($$\Delta a$$) is $$\Delta a = |a_1 - a_2|$$

$$\Delta a = (a_s + a_m) - (a_s - a_m)$$

$$\Delta a = 2 \cdot a_m$$

The change is simply twice the acceleration produced by the Moon on the Earth.

The acceleration produced by the Moon on the Earth is given by Newton's Law of Universal Gravitation:

$$a_m = \frac{G \cdot M_{moon}}{r^2}$$

$$a_m = \frac{6.6 \times 10^{-11} \cdot 7.36 \times 10^{22}}{(3.8 \times 10^8)^2}$$
$$a_m = \frac{48.576 \times 10^{11}}{14.44 \times 10^{16}}$$
$$a_m \approx 3.364 \times 10^{-5} \text{ m/s}^2$$

$$\Delta a = 2 \cdot a_m$$

$$\Delta a = 6.73 \times 10^{-5} \text{ m/s}^2$$

We have a planet of mass $$M$$ and radius $$R$$. A satellite of mass $$m$$ is to be placed in a circular orbit whose height above the planet’s surface is $$2R$$. Therefore the radius of the orbit, measured from the planet’s centre, is

$$r = R + 2R = 3R.$$

First, let us recall the expression for the gravitational potential energy of a mass $$m$$ at a distance $$r$$ from the centre of a planet of mass $$M$$:

$$U = -\dfrac{G M m}{r}.$$

Next, for a circular orbit we use the condition that the gravitational force provides the necessary centripetal force. Stating the formula,

$$\dfrac{G M m}{r^{2}} = \dfrac{m v^{2}}{r}.$$

Cancelling one factor of $$r$$ and solving for the orbital speed $$v$$ gives

$$v^{2} = \dfrac{G M}{r}, \qquad\text{so}\qquad v = \sqrt{\dfrac{G M}{r}}.$$

The kinetic energy of the satellite in this circular orbit is therefore

$$K = \dfrac{1}{2} m v^{2} = \dfrac{1}{2} m \left(\dfrac{G M}{r}\right) = \dfrac{G M m}{2 r}.$$

The total mechanical energy in the orbit is the sum of kinetic and potential energies:

$$E_{\text{orbit}} = K + U = \dfrac{G M m}{2 r} \;-\; \dfrac{G M m}{r} = -\dfrac{G M m}{2 r}.$$

Substituting $$r = 3R$$, we obtain

$$E_{\text{orbit}} = -\dfrac{G M m}{2 \times 3R} = -\dfrac{G M m}{6R}.$$

Now, let us look at the energy of the satellite initially, when it is resting on the planet’s surface. Because it starts from rest, its kinetic energy is zero, and its potential energy is

$$U_{\text{surface}} = -\dfrac{G M m}{R}.$$

Hence the initial total energy is simply

$$E_{\text{initial}} = -\dfrac{G M m}{R}.$$

The minimum energy that must be supplied by the launching mechanism equals the increase in the satellite’s total mechanical energy as it goes from the surface to the required orbit. Therefore,

$$\begin{aligned} \Delta E &= E_{\text{orbit}} - E_{\text{initial}} \\ &= \left(-\dfrac{G M m}{6R}\right) \;-\; \left(-\dfrac{G M m}{R}\right) \\ &= -\dfrac{G M m}{6R} + \dfrac{G M m}{R}. \end{aligned}$$

Bringing the terms to a common denominator of $$6R$$, we have

$$\Delta E = -\dfrac{G M m}{6R} + \dfrac{6 G M m}{6R} = \dfrac{5 G M m}{6R}.$$

This quantity is positive, meaning it represents the energy that must be provided. Thus the minimum launch energy is

$$\boxed{\dfrac{5 G M m}{6R}}.$$

Comparing with the given options, this matches Option C.

Hence, the correct answer is Option C.